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Full text of "Everything Maths: Grade 12 Mathematics"

VERSION 0.9 NCS 



GRADE 12 
MATHEMATICS 

WRITTEN BY VOLUNTEERS 



SlYAVULA 

TECHNOLOGY-POWERED LEARNING 



6 

■ 



I Trigonometry exercises in 
this book 

I Geometry exercises in this book 

I Algebra exercises in this book 




Litres of ink used in the 
production of alt the grade 10, 
11 and 12 textbooks 

Litres of gtue used in the 
production of all the gradelO, 
11 and 12 textbooks 













1 27 








■ 25 






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1342 








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teaching from this book 



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Maths textbook 

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Maths textbook 

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9 



basic education 

Department: 

Basic Education 

REPUBLIC OF SOUTH AFRICA 



* 



SHUTTLEWORTH 

FUNDED 



Everything Maths 



Grade 12 Mathematics 



Version 0.9- NCS 



by Siyavula and volunteers 



Copyright notice 



Your freedom to legally copy this book 

You are allowed and encouraged to freely copy this book. You can photocopy, print and 
distribute it as often as you like. You can download it onto your mobile phone, iPad, PC or 
flash drive. You can burn it to CD, e-mail it around or upload it to your website. 

The only restriction is that you have to keep this book, its cover and short-codes unchanged. 

For more information about the Creative Commons Attribution-NoDerivs 3.0 Unported (CC 
BY-ND 3.0) license see http://creativecommons.Org/licenses/by-nd/3.0/ 



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CC 



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Authors List 

This book is based upon the original Free High School Science Text which was entirely written 
by volunteer academics, educators and industry professionals. Their vision was to see a cur- 
riculum aligned set of mathematics and physical science textbooks which are freely available 
to anybody and exist under an open copyright license. 

Siyavula core team 

Neels van der Westhuizen; Alison Jenkin; Marina van Zyl; Helen Robertson; Carl Scheffler; Nicola du Toit; Leonard Gumani 
Mudau; 

Original Free High School Science Texts core team 

Mark Horner; Samuel Halliday; Sarah BIyth; Rory Adams; Spencer Wheaton 

Original Free High School Science Texts editors 

Jaynie Padayachee; Joanne Boulle; Diana Mulcahy; Annette Nell; Ren Toerien; Donovan Whitfield 

Siyavula and Free High School Science Texts contributors 

Sarah Abel; Dr. Rory Adams; Andrea Africa; Matthew Amundsen; Ben Anhalt; Prashant Arora; Amos Baloyi; Bongani Baloyi; 
Raymond Barbour; Caro-Joy Barendse; Richard Baxter; Tara Beckerling; Dr. Sarah BIyth; Sebastian Bodenstein; Martin Bongers; 
Gareth Boxall; Stephan Brandt; Hannes Breytenbach; Alex Briell; Wilbur Britz; Graeme Broster; Craig Brown; Richard Burge; 
Bianca Bhmer; George Calder-Potts; Eleanor Cameron; Richard Case; Sithembile Cele; Alice Chang; Richard Cheng; Fanny 
Cherblanc; Dr. Christine Chung; Brett Cocks; Stefaan Conradie; Rocco Coppejans; Tim Craib; Andrew Craig; Tim Crombie; 
Dan Crytser; Dr. Anne Dabrowski; Laura Daniels; Gareth Davies; Jennifer de Beyer; Jennifer de Beyer; Deanne de Bude; Mia 
de Vos; Sean Dobbs; Buhle Donga; William Donkin; Esmi Dreyer; Nicola du Toit; Matthew Duddy; Fernando Durrell; Dr. 
Dan Dwyer; Alex Ellis; Tom Ellis; Andrew Fisher; Giovanni Franzoni; Nina Gitau Muchunu; Lindsay Glesener; Kevin Godby; 
Dr. Vanessa Godfrey; Terence Goldberg; Dr. Johan Gonzalez; Saaligha Gool; Hemant Gopal; Dr. Stephanie Gould; Umeshree 
Govender; Heather Gray; Lynn Greeff; Carine Grobbelaar; Dr. Tom Gutierrez; Brooke Haag; Kate Hadley; Alex Hall; Dr. Sam 
Halliday; Asheena Hanuman; Dr. Nicholas Harrison; Neil Hart; Nicholas Hatcher; Jason Hayden; Laura Hayward; Cho Hee 
Shrader; Dr. Fritha Hennessy; Shaun Hewitson; Millie Hilgart; Grant Hillebrand; Nick Hobbs; Chris Holdsworth; Dr. Benne 
Holwerda; Dr. Mark Horner; Robert Hovden; Mfandaidza Hove; Jennifer Hsieh; Laura Huss; Dr. Matina J. Rassias; Rowan 
Jelley; Grant Jelley; Clare Johnson; Luke Jordan; Tana Joseph; Dr. Fabian Jutz; Brian Kamanzi; Dr. Lutz Kampmann; Simon 
Katende; Natalia Kavalenia; Nothando Khumalo; Paul Kim; Dr. Jennifer Klay; Lara Kruger; Sihle Kubheka; Andrew Kubik 
Dr. Jannie Leach; Nkoana Lebaka; Dr. Tom Leinster; Henry Liu; Christopher Loetscher; Mike Loseby; Amandla Mabona 
Malothe Mabutho; Stuart Macdonald; Dr. Anton Machacek; Tshepo Madisha; Batsirai Magunje; Dr. Komal Maheshwari 
Michael Malahe; Masoabi Malunga; Masilo Mapaila; Bryony Martin; Nicole Masureik; John Mathew; Dr. Will Matthews 
Chiedza Matuso; JoEllen McBride; Dr Melanie Dymond Harper; Nikolai Meures; Riana Meyer; Filippo Miatto; Jenny Miller; 
Abdul Mirza; Mapholo Modise; Carla Moerdyk; Tshwarelo Mohlala; Relebohile Molaoa; Marasi Monyau; Asogan Moodaly 
Jothi Moodley; Robert Moon; Calvin Moore; Bhavani Morarjee; Kholofelo Moyaba; Kate Murphy; Emmanuel Musonza; Tom 
Mutabazi; David Myburgh; Kamie Naidu; Nolene Naidu; Gokul Nair; Vafa Naraghi; Bridget Nash; Tyrone Negus; Huw 
Newton-Hill; Buntu Ngcebetsha; Dr. Markus Oldenburg; Thomas ODonnell; Dr. William P. Heal; Dr. Jaynie Padayachee; 
Poveshen Padayachee; Masimba Paradza; Dave Pawson; Justin Pead; Nicolette Pekeur; Sirika Pill ay; Jacques Plaut; Barry 
Povey; Barry Povey; Andrea Prinsloo; Joseph Raimondo; Sanya Rajani; Alastair Ramlakan; Dr. Jocelyn Read; Jonathan Reader; 
Jane Reddick; Dr. Matthew Reece; Razvan Remsing; Laura Richter; Max Richter; Sean Riddle; Dr. David Roberts; Christopher 
Roberts; Helen Robertson; Evan Robinson; Raoul Rontsch; Dr. Andrew Rose; Katie Ross; Jeanne-Mari Roux; Mark Roux; 
Bianca Ruddy; Nitin Rughoonauth; Katie Russell; Steven Sam; Dr. Carl Scheffler; Nathaniel Schwartz; Duncan Scott; Helen 
Seals; Relebohile Sefako; Prof. Sergey Rakityansky; Sandra Serumaga-Zake; Paul Shangase; Cameron Sharp; Ian Sherratt; Dr. 
James Short; Roger Sieloff; Brandon Sim; Bonga Skozana; Clare Slotow; Bradley Smith; Greg Solomon; Nicholas Spaull; Dr. 
Andrew Stacey; Dr. Jim Stasheff; Mike Stay; Mike Stringer; Masixole Swartbooi; Tshenolo Tau; Tim Teatro; Ben Tho.epson; 
Shen Tian; Xolani Timbile; Robert Torregrosa; Jimmy Tseng; Tim van Beek; Neels van der Westhuizen; Frans van Eeden; Pierre 
van Heerden; Dr. Marco van Leeuwen; Marina van Zyl; Pieter Vergeer; Rizmari Versfeld; Mfundo Vezi; Mpilonhle Vilakazi; 
Ingrid von Glehn; Tamara von Glehn; Kosma von Maltitz; Helen Waugh; Leandra Webb; Dr. Dawn Webber; Michelle Wen; 
Dr. Alexander Wetzler; Dr. Spencer Wheaton; Vivian White; Dr. Gerald Wigger; Harry Wiggins; Heather Williams; Wendy 
Williams; Julie Wilson; Timothy Wilson; Andrew Wood; Emma Wormauld; Dr. Sahal Yacoob; Jean Youssef; Ewald Zietsman 



Everything Maths 



Mathematics is commonly thought of as being about numbers but mathematics is actually a language! 
Mathematics is the language that nature speaks to us in. As we learn to understand and speak this lan- 
guage, we can discover many of nature's secrets. Just as understanding someone's language is necessary 
to learn more about them, mathematics is required to learn about all aspects of the world - whether it 
is physical sciences, life sciences or even finance and economics. 

The great writers and poets of the world have the ability to draw on words and put them together in ways 
that can tell beautiful or inspiring stories. In a similar way, one can draw on mathematics to explain and 
create new things. Many of the modern technologies that have enriched our lives are greatly dependent 
on mathematics. DVDs, Google searches, bank cards with PIN numbers are just some examples. And 
just as words were not created specifically to tell a story but their existence enabled stories to be told, so 
the mathematics used to create these technologies was not developed for its own sake, but was available 
to be drawn on when the time for its application was right. 

There is in fact not an area of life that is not affected by mathematics. Many of the most sought after 
careers depend on the use of mathematics. Civil engineers use mathematics to determine how to best 
design new structures; economists use mathematics to describe and predict how the economy will react 
to certain changes; investors use mathematics to price certain types of shares or calculate how risky 
particular investments are; software developers use mathematics for many of the algorithms (such as 
Google searches and data security) that make programmes useful. 

But, even in our daily lives mathematics is everywhere - in our use of distance, time and money. 
Mathematics is even present in art, design and music as it informs proportions and musical tones. The 
greater our ability to understand mathematics, the greater our ability to appreciate beauty and everything 
in nature. Far from being just a cold and abstract discipline, mathematics embodies logic, symmetry, 
harmony and technological progress. More than any other language, mathematics is everywhere and 
universal in its application. 

See introductory video by Dr. Mark Horner: VMiwd at www.everythingmaths.co.za 



More than a regular textbook 



Mobile version 



PC and tablet version 



Television broadcast 



3) *V 3) 



Video and rich media 









Practice and self test 



^^ WebBook I 

Summary presentations Post and see questions 




Everything Maths is not just a Mathematics textbook. It has everything you expect from your regular 
printed school textbook, but comes with a whole lot more. For a start, you can download or read it 
on-line on your mobile phone, computer or iPad, which means you have the convenience of accessing 
it wherever you are. 

We know that some things are hard to explain in words. That is why every chapter comes with video 
lessons and explanations which help bring the ideas and concepts to life. Summary presentations at 
the end of every chapter offer an overview of the content covered, with key points highlighted for easy 
revision. 

All the exercises inside the book link to a service where you can get more practice, see the full solution 
or test your skills level on mobile and PC. 

We are interested in what you think, wonder about or struggle with as you read through the book and 
attempt the exercises. That is why we made it possible for you to use your mobile phone or computer 
to digitally pin your question to a page and see what questions and answers other readers pinned up. 



Everything Maths on your mobile or PC 

You can have this textbook at hand wherever you are - whether at home, on the the train or at school. 
Just browse to the on-line version of Everything Maths on your mobile phone, tablet or computer. To 
read it off-line you can download a PDF or e-book version. 

To read or download it, go to www.everythingmaths.co.za on your phone or computer. 





Using the icons and short-codes 

Inside the book you will find these icons to help you spot where videos, presentations, practice tools 
and more help exist. The short-codes next to the icons allow you to navigate directly to the resources 
on-line without having to search for them. 



(A123) Go directly to a section 

(►j (V123) Video, simulation or presentation 

Cf\*j (P123) Practice and test your skills 

C?y (Q123) Ask for help or find an answer 



To watch the videos on-line, practise your skills or post a question, go to the Everything Maths website 
at www.everythingmaths.co.za on your mobile or PC and enter the short-code in the navigation box. 



Video lessons 



Look out for the video icons inside the book. These will take you to video lessons that help bring the 
ideas and concepts on the page to life. Get extra insight, detailed explanations and worked examples. 
See the concepts in action and hear real people talk about how they use maths and science in their 
work. 



See video explanation (►) (Video: V123) 





Video exercises 



Wherever there are exercises in the book you will see icons and short-codes for video solutions, practice 
and help. These short-codes will take you to video solutions of select exercises to show you step-by-step 
how to solve such problems. 



See video exercise (►) (Video: V123) 




You can get these videos by: 



• viewing them on-line on your mobile or computer 

• downloading the videos for off-line viewing on your phone or computer 

• ordering a DVD to play on your TV or computer 

• downloading them off-line over Bluetooth or Wi-Fi from select outlets 

To view, download, or for more information, visit the Everything Maths website on your phone or 
computer at www.everythingmaths.co.za 



Practice and test your skills 



One of the best ways to prepare for your tests and exams is to practice answering the same kind of 
questions you will be tested on. At every set of exercises you will see a practice icon and short-code. 
This on-line practice for mobile and PC will keep track of your performance and progress, give you 
feedback on areas which require more attention and suggest which sections or videos to look at. 



See more practice (A*) (QM123) 




To practice and test your skills: 

Go to www.everythingmaths.co.za on your mobile phone or PC and enter the short-code. 



Answers to your questions 



Have you ever had a question about a specific fact, formula or exercise in your textbook and wished 
you could just ask someone? Surely someone else in the country must have had the same question at 
the same place in the textbook. 



9:41 AM 






GRADE 10 
MATHEMATICS 


(2) Help : 006d 












Factorise : [2k 2 j + 24fc 2 j 2 








• Is 12, k and j common factors? | 

• Is it best to use k2j or kj2 as [ 
common factor? { 

• What does factorise ean? 








mathImat.cs ^^^ 06 ^ 



Factorise : 12k 2 j + 2\k 2 f 



Shud I devide out both k 
and j or just one? 



n 



P You should take out all the 
t - common factors, so 12, k 2 and 

j because they appear in both 

terms 




Database of questions and answers 

We invite you to browse our database of questions and answer for every sections and exercises in the 
book. Find the short-code for the section or exercise where you have a question and enter it into the 
short-code search box on the web or mobi-site at www.everythingmaths.co.za or 
www.everythingscience.co.za. You will be directed to all the questions previously asked and answered 
for that section or exercise. 



(A1 23) Visit this section to post or view questions 
C?J (Q123) Questions or help with a specific question 



Ask an expert 

Can't find your question or the answer to it in the questions database? Then we invite you to try our 
service where you can send your question directly to an expert who will reply with an answer. Again, 
use the short-code for the section or exercise in the book to identify your problem area. 



Contents 



1 Introduction to Book 

1.1 The Language of Mathematics 



2.6 Logarithm Law 2 

2.7 Logarithm Law 3 



2 Logarithms 4 

2.1 Introduction 4 

2.2 Definition of Logarithms 4 

2.3 Logarithm Bases 5 

2.4 Laws of Logarithms 6 

2.5 Logarithm Law 1 : log a 1 = 6 

log» = 1 7 

l°ga<>-2/) = log a (x) +log a (j/) 8 

2.8 Logarithm Law 4: log a (|J = log a (x) - log a (y) 8 

2.9 Logarithm Law 5: log a (x f ') = 61og a (x) 9 

2.10 Logarithm Law 6: log a (^x) = ^M 10 

2.11 Solving Simple Log Equations 12 

2.12 Logarithmic Applications in the Real World 15 

3 Sequences and Series 19 

3.1 Introduction 19 

3.2 Arithmetic Sequences 19 

3.3 Geometric Sequences 21 

3.4 Recursive Formulae for Sequences 25 



3.5 Series 



26 



3.6 Finite Arithmetic Series 28 

3.7 Finite Squared Series 31 

3.8 Finite Geometric Series 32 

3.9 Infinite Series 34 

4 Finance 39 

4.1 Introduction 39 

4.2 Finding the Length of the Investment or Loan 39 

4.3 Series of Payments 40 

4.4 Investments and Loans 48 

4.5 Formula Sheet 53 

5 Factorising Cubic Polynomials 56 

5.1 Introduction 56 

5.2 The Factor Theorem 56 

5.3 Factorisation of Cubic Polynomials 58 

5.4 Solving Cubic Equations 61 



11 



CONTENTS CONTENTS 

6 Functions and Graphs 65 

6.1 Introduction 65 

6.2 Definition of a Function 65 

6.3 Notation Used for Functions 66 

6.4 Graphs of Inverse Functions 67 

7 Differential Calculus 75 

7.1 Introduction 75 

7.2 Limits 75 

7.3 Differentiation from First Principles 86 

7.4 Rules of Differentiation 88 

7.5 Applying Differentiation to Draw Graphs 90 

7.6 Using Differential Calculus to Solve Problems 99 

8 Linear Programming 108 

8.1 Introduction 108 

8.2 Terminology 108 

8.3 Linear Programming and the Feasible Region 109 

9 Geometry 117 

9.1 Introduction 117 

9.2 Circle Geometry 117 

9.3 Co-ordinate Geometry 138 

9.4 Transformations 143 

10 Trigonometry 148 

10.1 Introduction 148 

10.2 Compound Angle Identities 148 

10.3 Applications of Trigonometric Functions 155 

10.4 Other Geometries 160 

11 Statistics 164 

11.1 Introduction 164 

11.2 Normal Distribution 164 

11.3 Extracting a Sample Population 166 

11.4 Function Fitting and Regression Analysis 167 

12 Combinations and Permutations 178 

12.1 Introduction 178 

12.2 Counting 178 

12.3 Notation 179 

12.4 Fundamental Counting Principle 179 

12.5 Combinations 180 

12.6 Permutations 182 

12.7 Applications 184 



Introduction to Book 





The Language of Mathematics 



The purpose of any language, like English or Zulu, is to make it possible for people to communicate. 
All languages have an alphabet, which is a group of letters that are used to make up words. There are 
also rules of grammar which explain how words are supposed to be used to build up sentences. This 
is needed because when a sentence is written, the person reading the sentence understands exactly 
what the writer is trying to explain. Punctuation marks (like a full stop or a comma) are used to further 
clarify what is written. 

Mathematics is a language, specifically it is the language of Science. Like any language, mathematics 
has letters (known as numbers) that are used to make up words (known as expressions), and sentences 
(known as equations). The punctuation marks of mathematics are the different signs and symbols that 
are used, for example, the plus sign (+), the minus sign (— ), the multiplication sign (x), the equals sign 
(=) and so on. There are also rules that explain how the numbers should be used together with the 
signs to make up equations that express some meaning. 

© See introductory video: VMinh at www.everythingmaths.co.za 



Logarithms 





2. 1 Introduction 




In mathematics many ideas are related. We saw that addition and subtraction are related and that 
multiplication and division are related. Similarly, exponentials and logarithms are related. 

Logarithms, commonly referred to as logs, are the inverse of exponentials. The logarithm of a number 
x in the base a is defined as the number n such that a n = x. 



So, if a" = x, then: 

log (x) = n 

© See introductory video: VMgio at www.everythingmaths.co.za 



(2.1) 




2.2 Definition of Logarithms 




The logarithm of a number is the value to which the base must be raised to give that number i.e. the 
exponent. From the first example of the activity log 2 (4) means the power of 2 that will give 4. As 
2 2 = 4, we see that 

log 2 (4) = 2 (2.2) 

The exponential-form is then 2 2 = 4 and the logarithmic-form is log 2 4 = 2. 



DEFINITION: Logarithms 



If a" = x, then: log a (:r) = n, where a > 0; a ^ 1 and x > 0. 



Activity. 



Logarithm Symbols 



Write the following out in words. The first one is done for you. 
1 . log 2 (4) islogtothebase2of4 

2- log 10 (14) 

3- log 16 (4) 

4. log, (8) 

5. log,, (a;) 



CHAPTER 2. LOGARITHMS 



2.3 



Activity: 



Applying the definition 



Find the value of: 
1. log 7 343 



2. log 2 8 
3- log 4 i 
4. log 10 1 000 



Reasoning : 

7 3 = 343 

therefore, log 7 343 = 3 




2.3 Logarithm Bases 




Logarithms, like exponentials, also have a base and log 2 (2) is not the same as log 10 (2). 

We generally use the "common" base, 10, or the natural base, e. 

The number e is an irrational number between 2.71 and 2.72. It comes up surprisingly often in Math- 
ematics, but for now suffice it to say that it is one of the two common bases. 



Extension: 



Natural Logarithm 



The natural logarithm (symbol In) is widely used in the sciences. The natural logarithm is to 
the base e which is approximately 2.71828183 .... e, like tt and is an example of an irrational 
number. 



While the notation log 10 (a;) and log e (x) may be used, log 10 (x) is often written log(x) in Science and 
log e (x) is normally written as ln(x) in both Science and Mathematics. So, if you see the log symbol 
without a base, it means log 10 . 

It is often necessary or convenient to convert a log from one base to another. An engineer might need 
an approximate solution to a log in a base for which he does not have a table or calculator function, 
or it may be algebraically convenient to have two logs in the same base. 

Logarithms can be changed from one base to another, by using the change of base formula: 

logtz 



log„ x 



log;, a 



(2.3) 



where b is any base you find convenient. Normally a and 6 are known, therefore log,, a is normally a 
known, if irrational, number. 



For example, change log 2 12 in base 10 is: 



log, 12 



logic 12 

lo Sio 2 



2.4 



CHAPTER 2. LOGARITHMS 



Activity: 



Change of Base 



Change the following to the indicated base: 

1 . log 2 (4) to base 8 

2. log 1(1 (14) to base 2 

3. log 16 (4) to base 10 

4. log x (8) to base y 

5. log (x) to base x 

© See video: VMgiq at www.everythingmaths.co.za 




2.4 Laws of Logarithms 




Just as for the exponents, logarithms have some laws which make working with them easier. These 
laws are based on the exponential laws and are summarised first and then explained in detail. 






log„(l) = 


= 


(2.4) 




i°g Q 0) = 


= 1 


(2.5) 




lo ga ( x ■ v) ~- 


= l°ga(z)+l°ga(2/) 


(2.6) 




*■(;) ■ 


= l°gaO) -l°g„ (2/) 


(2.7) 




l°g„ (^J = 


= &log a (a:) 


(2.8) 




log a (^) = 


l0g o (!c) 

b 


(2.9) 



2.5 Logarithm Law 1: log a 1 = 






c- 

Since a 




Then, log a (l) 


For example, 




and 





1 

by definition of logarithm in Equation 2.1 

log 2 1 = 
log,, 1 = 



Activity: 



Logarithm Law 7: log a 1 = 



CHAPTER 2. LOGARITHMS 



2.6 



Simplify the following: 

1. log 2 (l) + 5 

2. log 10 (l) x 100 

3. 3 x log 16 (l) 

4. \og x (l) + 2xy 

c l°Sv(l) 




2.6 Logarithm Law 2: log a (a 




For example, 



and 



r- 1 

Since a = a 
Then, log a (a) = 1 by definition of logarithm in Equation 2.1 



log, 2 = 1 



log 25 25 = 1 



Activity: 



Logarithm Law 2: log„(a) = 1 



Simplify the following: 

1. log 2 (2) + 5 

2. log 10 (10) x 100 

3. 3 x log 16 (16) 

4. logj,(a;) + 2xy 



5. 



l°g„(K) 



Tip 



When the base is 10 we do not need to state it. From the work done up to now, it is also useful to 
summarise the following facts: 



Useful to know and re- 
member 



1 . log 1 = 

2. log 10 = 1 

3. log 100 = 2 

4. log 1000 = 3 



2.7 



CHAPTER 2. LOGARITHMS 




2.7 Logarithm Law 3: 

log a (x-2/) = log a (x) + log a (y) 




The derivation of this law is a bit trickier than the first two. Firstly, we need to relate x and y to the 
base a. So, assume that x = a m and y = a". Then from Equation 2.1, we have that: 

log a (x) = m (2.10) 

and logjy) = n (2.11) 

This means that we can write: 

log a (»•!/) = log a (a m .a") 

= log a (a m+ ") (Exponential Law Equation (Grade 10)) 

= log a (a log " 0E)+log " (: " ) ) (From Equation 2.10 and Equation 2.1 1) 

= log (x) + log a (y) (From Equation 2.1) 

For example, show that log(10 . 100) = log 10 + log 100. Start with calculating the left hand side: 

log(10.100) = log(1000) 
= log(10 3 ) 
= 3 

The right hand side: 

log 10 + log 100 = 1 + 2 
= 3 
Both sides are equal. Therefore, log(10 . 100) = log 10 + log 100. 



Activity. 



Logarithm Law 3: log Q (x . y) = log (jc) + log (y) 



Write as separate logs: 

1. log 2 (8x4) 

2. log 8 (10 x 10) 
3- log l6 (xy) 

4. \og z {2xy) 
5- log, fe 2 ) 



2.8 Logarithm Law 4: 

lQ ga (J) = lQ ga(^) - l0g a (2/) 



EMCI 



The derivation of this law is identical to the derivation of Logarithm Law 3 and is left as an exercise. 
For example, show that log(^jjj) = log 10 — log 100. Start with calculating the left hand side: 

1 

logClO- 1 ) 
-1 



logf^) 

s \100J 



ll « 



CHAPTER 2. LOGARITHMS 



2.9 



The right hand side: 



log 10 - log 100 = 1-2 
= -1 



Both sides are equal. Therefore, log(j^) = log 10 — log 100. 



Activity: 



Logarithm Law 4: log a ( | J = log a (a;) - log„(y) 



Write as separate logs: 
1. log 2 (|) 

2- log 8 (^) 

3- log 16 (f) 

4- log 2 (f) 

5- log x (|) 



2.9 Logarithm Law 5: 

lo ga(^) = bhg a {x 




Once again, we need to relate x to the base a. So, we let x = a m . Then, 

loga^') = l°ga (( a m ) b ) 

= log a (a m ' ) (Exponential Law in Equation (Grade 10)) 

But, m = log a (x) (Assumption that x = a m ) 

.; logjz 6 ) = ]og„(a'- to «- W ) 

= b.log a (x) (Definition of logarithm in Equation 2.1) 

For example, we can show that log 2 (5 3 ) = 31og 2 (5). 
log 2 (5 3 ) = log 2 (5.5.5) 



log 2 5 + log 2 5 + log 2 5 (■.■ log a (:r . y) = log Q (a . a )) 
3 log, 5 



Therefore, log 2 (5 3 ) = 31og 2 (5). 



Activity: 



Logarithm Law 5: log a (z 6 ) = blog a (x) 



Simplify the following: 

1. log 2 (8 4 ) 

2. log 8 (10 10 ) 

3. log 16 (x") 

4- iog z (y x ) 

5- log,(2/ 2x ) 



2.10 



CHAPTER 2. LOGARITHMS 




2.10 Logarithm Law 6: 




The derivation of this law is identical to the derivation of Logarithm Law 5 and is left as an exercise. 
For example, we can show that log 2 (-v / 5) = ° s 3 . 



log 2 (^5) = log 2 (5*) 
1 
3 
log 2 5 



log 2 5 (■.' log^a/") = 61og a (:r)) 



Therefore, log^v^) = l£ f- 5 . 



Activity: 



Logarithm Law 6: log a {J/x) 



logq(^) 



Tip 



The final answer doesn't 
have to look simple. 



Simplify the following: 

1. log 2 (^8) 

2. log 8 ( Vld) 

3- log 16 (^/z) 

4- log 2 (^) 

® See video: VMgjI at www.everythingmaths.co.za 



Example 1: Simplification of Logs 



QUESTION 


Simplify, without use of a 


calculator: 


3 log 3 + log 125 


SOLUTION 






Step 1 : Try to write any quantities as exponents 

125 can be written as 5 3 . 


Step 2 : Simplify 







10 



CHAPTER 2. LOGARITHMS 2.10 

3 log 3 + log 125 = 3 log 3 + log 5 3 

= 31og3 + 31og5 ■ : \og a {x b ) = b\og a {x) 
= 3 log 15 (Logarithm Law 3) 
Step 3 : Final Answer 

We cannot simplify any further. The final answer is: 

3 log 15 



Example 2: Simplification of Logs 



QUESTION 



Simplify, without use of a calculator: 

8^ +log 2 32 



SOLUTION 



Step 1 : Try to write any quantities as exponents 

8 can be written as 2 3 . 32 can be written as 2 5 . 

Step 2 : Re-write the question using the exponential forms of the numbers 

8* +log 2 32 = (2 3 )i +log 2 2 5 

Step 3 : Determine which laws can be used. 

We can use: 

Step 4 : Apply log laws to simplify 

(2 3 )I+log 2 2 5 = (2) 3x I+51og 2 2 

Step 5 : Determine which laws can be used. 

We can now use log a a = 1 

Step 6 : Apply log laws to simplify 

(2) 2 + 5 log 2 2 = 2 2 + 5(1) = 4 + 5 = 9 

Step 7 : Final Answer 

The final answer is: 

83 +log 2 32 = 9 



11 



2.11 



CHAPTER 2. LOGARITHMS 



Example 3: Simplify to one log 




QUESTION 


Write 2 log 3 + log 2 - log 5 


as the logarithm 


of a singl 


e number. 


SOLUTIOS 












Step 7 : 


Reverse law 5 

2 log 3 + log 2 


- log 5 = 


log3 2 


+ log2- 


log 5 


Step 2 : 


Apply laws 3 and 4 

= log(3 2 x 2 -=- 5) 








Step 3 : 


Write the final 

= log 3,6 


answer 











2. / / Solving Simple Log Equations 




In Grade 10 you solved some exponential equations by trial and error, because you did not know the 
great power of logarithms yet. Now it is much easier to solve these equations by using logarithms. 

For example to solve x in 25 x = 50 correct to two decimal places you simply apply the following 
reasoning. IftheLHS = RHS then the logarithm of the LHS must be equal to the logarithm of the RHS. 
By applying Law 5, you will be able to use your calculator to solve for x. 



Example 4: Solving Log equations 



QUESTION 



Solve for x: 25* = 50 correct to two decimal places. 



12 



CHAPTER 2. LOGARITHMS 2.11 



SOLUTION 



Step 1 : Taking the log of both sides 

log 25 x = log 50 

Step 2 : Use Law 5 

x log 25 = log 50 

Step 3 : Solve for x 

x = log 50 — log 25 
x = 1,21533... 

Step 4 : Round off to required decimal place 

x = 1,22 



In general, the exponential equation should be simplified as much as possible. Then the aim is to 
make the unknown quantity (i.e. x) the subject of the equation. 

For example, the equation 

is solved by moving all terms with the unknown to one side of the equation and taking all constants to 
the other side of the equation 

2 X . 2 2 = 1 
1 

2 = 2^ 

Then, take the logarithm of each side. 

log (2*) = log(l 

slog (2) = -log(2 2 ) 

x log (2) = -2 log (2) Divide both sides by log (2) 
.'. x = -2 



2+2 = 2° = 1 / 



Substituting into the original equation, yields 
Similarly, 9 (1 ~ 2a,) = 3 4 is solved as follows: 

g(l-2x) _ g4 

o2(l-2i) n4 

32-4* _ 3 4 take the | g ar j tnm f both sides 

log(3 2 - 4 *) = log(3 4 ) 

(2 - ix) log(3) = 41og(3) divide both sides by log(3) 

2 - Ax = 4 

-Ax = 2 

1 

:.x = -- 

Substituting into the original equation, yields 

g(l-2(^l)) _ g(l + l) _ 3 2(2) _ g4 j 



13 



2.11 CHAPTER 2. LOGARITHMS 



Example 5 


: Exponential Equation 




QUESTION 


Solve for x ir 


7 , 5 (3*+3) =35 




SOLUTION 






Step 1 : 


Identify the base with x as an exponent 

There are two possible bases: 5 and 7. x is an exponent of 5. 




Step 2 : 


Eliminate the base with no x 

In order to eliminate 7, divide both sides of the equation by 7 to 

5 (3*+3) = 5 


give: 


Step 3 : 


7a/re the logarithm of both sides 

log(5 (te+3) ) = log(5) 




Step 4 : 


Apply the log laws to make x the subject of the equation. 




(3x + 3) log(5) = log(5) divide both sides of the equation by 


log(5) 




3x + 3 = 1 






Zx = -2 

2 

X = —3 




Step 5 : 


Substitute into the original equation to check answer. 

7 .5«-3x|> + 3) =7 _ 5 (-2 + 3) =7 5 1 =35 / 







Exercise 2-1 



Solve for x: 

1 . log 3 x = 2 

2. I0'°s 27 = x 

3. 3 2x - 1 = 2i lx - 1 



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(1.)01bn (2.)01bp (3.) 01 bq 



It 



CHAPTER 2. LOGARITHMS 



2.12 




2.12 Logarithmic Applications in 
the Real World 



Logarithms are part of a number of formulae used in the Physical Sciences. There are formulae that 
deal with earthquakes, with sound, and pH-levels to mention a few. To work out time periods is growth 
or decay, logs are used to solve the particular equation. 



Example 6: Using the growth formula 



QUESTION 


A city grows 5% every two years. 


How long will it take for the city to triple its size? 


SOLUTION 






Step 1 


Use the formula 








A = P(l + i) n As 


ume P = x, then A = 3x. For this example 


n represents a 




period of 2 years, therefore the n is halved for this purpose. 




Step 2 


Substitute information given into formula 






3 


= (1,05)5 






log 3 


= — x log 1,05 (using Law 5) 






n 


= 2 log 3 H- log 1,05 






n 


= 45,034 




Step 3 


Final answer 








So it will take approximately 45 years for the population to triple 


in size. 





Exercise 2-2 



1. The population of a certain bacteria is expected to grow exponentially at a rate of 15% every 
hour. If the initial population is 5 000, how long will it take for the population to reach 100 000? 



2. Plus Bank is offering a savings account with an interest rate if 10% per annum compounded 
monthly. You can afford to save R300 per month. How long will it take you to save R20 000? 
(Give your answer in years and months) 



2A2 



CHAPTER 2. LOGARITHMS 



A"j More practice Crj video solutions Cf) or help at www.everythingmaths.co.za 



(l.)Olbr (2.)01bs 



Example 7: Logs in Compound Interest 




QUESTION 


1 have R12 000 to invest. 1 need the money 


to grow to at least R30 000. If it is invested at a 


compound interest rate of 13% per annum, 


for how long (in full years) does my investment 


need to grow? 




SOLUTION 




Step 1 : The formula to use 




A = 


P(l + i) n 


Step 2 : Substitute and solve for n 




30000 < 


12000(1 + 0,13)" 


1,13" > 


5 
2 


nlog(l,13) > 


log(2,5) 


n > 


log(2,5)-log(l,13) 


n > 


7,4972 . . . 


Step 3 : Determine the final answer 




In this case we round up, because 7 years will not yet deliver the required 


R30 000. The investment need to stay in the bank for at least 8 years. 





Chapter 2 


End of Chapter Exercises 



1 . Show that 



lo Sa 



;.W -i°g„(y) 



16 



CHAPTER 2. LOGARITHMS 2.12 



2. Show that 

io Sa m = ^ 

3. Without using a calculator show that: 

, 75 „. 5 . 32 , „ 

log 2 toe h log = log 2 

6 16 6 9 & 243 & 

4. Given that 5 n = x and n = log 2 y 

(a) Write y in terms of n 

(b) Express log 8 4y in terms of n 

(c) Express 50" +1 in terms of x and y 

5. Simplify, without the use of a calculator: 

(a) 8 1 +log 2 32 

(b) log 3 9 - log 5 V5 

1 

^(ft4ft)' +1 ^ 92 ' 12 

6. Simplify to a single number, without use of a calculator: 

,„„ log 32 -log 8 

(a) log 5 125 + -2- — — S- 

log8 

(b) log 3 -log 0,3 

7. Given: log 3 6 = a and log 6 5 = b 

(a) Express log 3 2 in terms of a. 

(b) Hence, or otherwise, find log 3 10 in terms of a and b. 

8. Given: pq k = qp^ 1 

Prove: k = 1 — 2 log g p 

9. Evaluate without using a calculator: (log 7 49) 5 + log 5 I — - ) — 13 log 9 1 

10. If log 5 = 0,7, determine, without using a calculator: 

(a) log 2 5 

(b) 10- 1 ' 4 

11. Given: M = log 2 (x + 3) + log 2 (x - 3) 

(a) Determine the values of x for which M is defined. 

(b) Solve for x if M = 4. 

/ \logx 

12. Solve: ( x a ) = I0x 2 (Answer(s) may be left in surd form, if necessary.) 



13. Find the value of (log 27 3) 3 without the use of a calculator. 

14. Simplify By using a calculator: log 4 8 + 21og 3 \/27 

15. Write log 4500 in terms of a and b if 2 = 10" and 9 = 10 6 . 

5 2006 _ 5 2004 + 24 



16. Calculate: 



1 



1 7. Solve the following equation for x without the use of a calculator and using the fact 
that\/lOP» 3,16 : 

21og(x + l)= „, 6 , t v -1 

log(x + 1) 

18. Solve the following equation for x: G bx = 66 (Give answer correct to two decimal 
places.) 



17 



2.12 CHAPTER 2. LOGARITHMS 



More practice Crj video solutions f 9) or help at www.everythingmaths.co.za 



(1.) Olbt (2.)01bu (3.)01bv (4.) 01 bw (5.) 01 bx (6.) 01 by 
(7.)01bz (8.)01c0 (9.)01c1 (10.)01c2 (1 1.) 01 c3 (12.)01c4 
(13.)01c5 (14.) 01c6 (15.)01c7 (16.)01c8 (17.)01c9 (18.) Olca 



is 



Sequences and Series 





3. 1 Introduction 




In this chapter we extend the arithmetic and quadratic sequences studied in earlier grades, to geometric 
sequences. We also look at series, which are the summing of the terms in sequences. 

© See introductory video: VMgjm at www.everythingmaths.co.za 




3.2 Arithmetic Sequences 



The simplest type of numerical sequence is an arithmetic sequence. 



DEFINITION: Arithmetic Sequence 

An arithmetic (or linear) sequence is a sequence of numbers in which each new term 
is calculated by adding a constant value to the previous term 



For example, 

1; 2; 3; 4; 5; 6; . . . 

is an arithmetic sequence because you add 1 to the current term to get the next term: 



first term 

second term 

third term 



2 = 1 + 1 

3 = 2+1 



n th term: n = (n - 1) + 1 



Activity: 



Common Difference 



Find the constant value that is added to get the following sequences and write out the next 
5 terms. 

1. 2; 6; 10; 14; 18; 22;... 

2. -5; -3; - 1; 1; 3; . . . 

3. 1; 4; 7; 10; 13; 16;... 

4. -1; 10; 21; 32; 43; 54;... 



19 



3.2 



CHAPTER 3. SEQUENCES AND SERIES 



5. 3; 0; -3; -6; -9; -12; 



General Equation for the n th -Term of an 
Arithmetic Sequence 



EMCP 



More formally, the number we start out with is called ai (the first term), and the difference between 
each successive term is denoted d, called the common difference. 

The general arithmetic sequence looks like: 



ai 


= 


fli 


a 2 


= 


Qi + d 


03 


= 


0,2 + d = (qi + d) + d = a\ + 2d 


cu 


= 


a-i + d = (qi + 2d) + d = a\ + 3d 


a n 


= 


ai + d . (n — 1) 



Thus, the equation for the ri"'-term is: 



a„ = ai + d. (n — 1) 



(3.1) 



Given ai and the common difference, d, the entire set of numbers belonging to an arithmetic sequence 
can be generated. 



Tip 



Test for Arithmetic Se- 
quences 



DEFINITION: Arithmetic Sequence 

An arithmetic (or linear) sequence is a sequence of numbers in which each new term 
is calculated by adding a constant value to the previous term: 



a„ = a„_i + d 



(3.2) 



where 



a„ represents the new term, the n" l -term, that is calculated; 
a„_i represents the previous term, the (n — l)"*-term; 
d represents some constant. 



A simple test for an arithmetic sequence is to check that the difference between consecutive terms is 
constant: 



ai — ai = a,3 — a 2 = a n — a n -\ = d 



(3.3) 



This is quite an important equation, and is the definitive test for an arithmetic sequence. If this condi- 
tion does not hold, the sequence is not an arithmetic sequence. 



2ll 



CHAPTER 3. SEQUENCES AND SERIES 



3.3 



Extension: 



Plotting a graph of terms in an arithmetic sequence 



Plotting a graph of the terms of a sequence sometimes helps in determining the type of se- 
quence involved. For an arithmetic sequence, plotting o„ vs. n results in: 



a n = Qi + d(n — 1) 




(Note that the graph will only be a straight line if the sequence is arithmetic.) 




3.3 Geometric Sequences 




DEFINITION: Geometric Sequences 

A geometric sequence is a sequence of numbers in which each new term (except for 
the first term) is calculated by multiplying the previous term by a constant value. 



This means that the ratio between consecutive numbers in the geometric sequence is a constant. We 
will explain what we mean by ratio after looking at the following example. 



21 



3.3 



CHAPTER 3. SEQUENCES AND SERIES 



Example - A Flu Epidemic 



EMCR 



Extension: 



What is influenza? 



Influenza (commonly called "flu") is caused by the influenza virus, which infects the respiratory 
tract (nose, throat, lungs). It can cause mild to severe illness that most of us get during winter 
time. The main way that the influenza virus is spread is from person to person in respiratory 
droplets of coughs and sneezes. (This is called "droplet spread".) This can happen when 
droplets from a cough or sneeze of an infected person are propelled (generally, up to a metre) 
through the air and deposited on the mouth or nose of people nearby. It is good practise to 
cover your mouth when you cough or sneeze so as not to infect others around you when you 
have the flu. 



Assume that you have the flu virus, and you forgot to cover your mouth when two friends came to visit 
while you were sick in bed. They leave, and the next day they also have the flu. Let's assume that they 
in turn spread the virus to two of their friends by the same droplet spread the following day. Assuming 
this pattern continues and each sick person infects 2 other friends, we can represent these events in 
the following manner: 




A A A. A. 

ijjiijjiffijjiijjiijpijji 



Figure 3.1 : Each person infects two more people with the flu virus. 



Again we can tabulate the events and formulate an equation for the general case: 



Day, n 


Number of newly-infected people 


1 


2=2 


2 


4 =2x2 = 2x2' 


3 


8 =2x4 = 2x2x2 = 2x2' 


4 


16 = 2x8 = 2x2x2x2 = 2x2^ 


5 


32 = 2 x 16 = 2 x 2 x 2 x 2 x 2 = 2 x 2 4 






n 


= 2x2x2x2x---x2 = 2x 2"~ x 



The above table represents the number of newly-infected people after n days since you first infected 
your 2 friends. 



22 



CHAPTER 3. SEQUENCES AND SERIES 



3.3 



You sneeze and the virus is carried over to 2 people who start the chain (on = 2). The next day, each 
one then infects 2 of their friends. Now 4 people are newly-infected. Each of them infects 2 people the 
third day, and 8 people are infected, and so on. These events can be written as a geometric sequence: 

2; 4; 8; 16; 32; ... 



Note the common ratio (2) between the events. Recall from the linear arithmetic sequence how the 
common difference between terms was established. In the geometric sequence we can determine the 
common ratio, r, from 



a > 



0.2 



Or, more generally, 



(3.4) 



(3.5) 



Activity: 



Common Ratio of Geometric Sequence 



Determine the common ratio for the following geometric sequences: 

1. 5; 10; 20; 40; 80;... 

i I- I- I- 

Z - 2> 4' 8' ' " 

3. 7; 28; 112; 448;... 

4. 2; 6; 18; 54;... 

5. -3; 30; - 300; 3000; . . . 



General Equation for the n th -Term of a Ge- 
ometric Sequence 



EMCS 



From the flu example above we know that ai = 2 and r = 2, and we have seen from the table that the 
n"*-term is given by a„ = 2 x 2 n_1 . Thus, in general, 



a„ = ai . r 



(3.6) 



where ai is the first term and r is called the common ratio. 

So, if we want to know how many people are newly-infected after 1 days, we need to work out aw: 

n-l 
a„ = a\ .r 

a w = 2 x 2 10 - 1 

= 2 x 2 9 

= 2 x 512 

= 1024 

That is, after 10 days, there are 1 024 newly-infected people. 



23 



3.3 



CHAPTER 3. SEQUENCES AND SERIES 



Or, after how many days would 16 384 people be newly-infected with the flu virus? 

n-l 



a„ 


= 


a\ . r 


16 384 


= 


2 x 2'- 1 


384-^2 


= 


2 „-i 


8 192 


= 


2 n-l 


2 13 


= 


2 „-l 


13 


= 


71- 1 


n 


= 


11 



That is, after 14 days 16 384 people are newly-infected. 



Activity: 



General Equation of Geometric Sequence 



Determine the formula for the n"'-term of the following geometric sequences: 

1. 5; 10; 20; 40; 80;... 

7 I- I- I- 

*■< 2 ' 4' 8' ' ' ' 

3. 7; 28; 112; 448; . . . 

4. 2,6,18,54,... 

5. -3; 30; - 300; 3000; . . . 



Exercise 3-1 



1 . What is the important characteristic of an arithmetic sequence? 

2. Write down how you would go about finding the formula for the n th term of an arithmetic 
sequence? 

3. A single square is made from 4 matchsticks. To make two squares in a row takes 7 matchsticks, 
while three squares in a row takes 10 matchsticks. Determine: 

(a) the first term 

(b) the common difference 

(c) the formula for the general term 

(d) how many matchsticks are in a row of 25 squares 




4. 5; x; y is an arithmetic sequence and x; y; 81 is a geometric sequence. All terms in the 
sequences are integers. Calculate the values of x and y. 



(A" 1 ) More practice (►) video solutions C 7} or help at www.everythingmaths.co.za 



(1.)01cb (2.)01cc (3.)01cd (4.) 01ce 



21 



CHAPTER 3. SEQUENCES AND SERIES 



3.4 




3.4 Recursive Formulae for 
Sequences 



When discussing arithmetic and quadratic sequences, we noticed that the difference between two 
consecutive terms in the sequence could be written in a general way. 

For an arithmetic sequence, where a new term is calculated by taking the previous term and adding a 
constant value, d: 

a„ = a n -i + d 

The above equation is an example of a recursive equation since we can calculate the n'^-term only by 
considering the previous term in the sequence. Compare this with Equation (3.1), 



a„ = oi + d. (n — 1) 



(3.7) 



where one can directly calculate the n"'-term of an arithmetic sequence without knowing previous 
terms. 

For quadratic sequences, we noticed the difference between consecutive terms is given by (??): 

a„ - a„_i = D . (n — 2) + d 

Therefore, we re-write the equation as 

o„ = On-i + D . (n - 2) + d (3.8) 

which is then a recursive equation for a quadratic sequence with common second difference, D. 
Using (3.5), the recursive equation for a geometric sequence is: 

a„ = r . a.n-i (3.9) 



Recursive equations are extremely powerful: you can work out every term in the series just by knowing 
previous terms. As you can see from the examples above, working out a n using the previous term a„_i 
can be a much simpler computation than working out a„ from scratch using a general formula. This 
means that using a recursive formula when using a computer to work out a sequence would mean the 
computer would finish its calculations significantly quicker. 



Activity: 



Recursive Formula 



Write the first five terms of the following sequences, given their recursive formulae: 

1 . a n = 2a n -i + 3, cl\ = 1 

2. a n = On— i,cn — 11 

3. a n = 2a n _ 1 ,ai = 2 



Extension: 



The Fibonacci Sequence 



Consider the following sequence: 

0; 1; 1; 2; 3; 5; 8; 13; 21; 34; 



(3.10) 



The above sequence is called the Fibonacci sequence. Each new term is calculated by adding 



3.5 



CHAPTER 3. SEQUENCES AND SERIES 



the previous two terms. Hence, we can write down the recursive equation: 

a„ = a„-i + a n -2 (3.11) 




3.5 Series 




In this section we simply work on the concept of adding up the numbers belonging to arithmetic and 
geometric sequences. We call the sum of any sequence of numbers a series. 



Some Basics 



EMCV 



If we add up the terms of a sequence, we obtain what is called a series. If we only sum a finite number 
of terms, we get a fin ite series. We use the symbol S„ to mean the sum of the first n terms of a sequence 
{ai; a2; 03; . . . ;a n }: 



On 



ai + 0.2 + a 3 + ■ ■ ■ + a„ 



(3.12) 



For example, if we have the following sequence of numbers 

1; 4; 9; 25; 36; 49; ... 
and we wish to find the sum of the first four terms, then we write 

Si = 1 + 4 + 9 + 25 = 39 
The above is an example of a finite series since we are only summing four terms. 
If we sum infinitely many terms of a sequence, we get an infinite series: 

Son = ai + 0-2 + a 3 + . . . 



(3.13) 



Sigma Notation 



EMCW 



In this section we introduce a notation that will make our lives a little easier. 

A sum may be written out using the summation symbol J^ • This symbol is sigma, which is the capital 
letter "S" in the Greek alphabet. It indicates that you must sum the expression to the right of it: 



Y.°- 



Om + Om+1 + ■ ■ ■ + Un-1 + O n 



(3.14) 



where 



is the index of the sum; 



2(i 



CHAPTER 3. SEQUENCES AND SERIES 3.5 

• to is the lower bound (or start index), shown below the summation symbol; 

• n is the upper bound (or end index), shown above the summation symbol; 

• a; is a term of a sequence. 

The index i increases from m to n in steps of 1. 

If we are summing from i = 1 (which implies summing from the first term in a sequence), then we can 
use either S„- or ^-notation since they mean the same thing: 

n 

S„ = ^^ o,i = 0.1 + a-2 + ■ ■ ■ + a n (3.15) 

For example, in the following sum, 

5 

£* 

we have to add together all the terms in the sequence a % = i from i = 1 up until i = 5: 

5 

^j= 1 + 2 + 3 + 4 + 5 = 15 

i i 

Examples 

1. 



£* 



2 + 4 + 8 + 16 + 32 + 64 
126 



2. 



J2(3x l ) = 3x 3 + 3x 4 + ■ ■ ■ + 3x g + 3x 



for any value x. 



Some Basic Rules for Sigma Notation 

1 . Given two sequences, a, and 6,, 



n n 



Y^{a t + bi) = Y. a ' + 12 bi (3 - 16) 

i=l i=l t=l 

2. For any constant c that is not dependent on the index i, 

n 

> c . a, = c . ai + c . a2 + c . a,s + ■ ■ ■ + c . a n 

,. i 

= c (ai + a 2 + 03 + ■ ■ ■ + a„) 

n 

= c^a, (3.17) 



27 



3.6 



CHAPTER 3. SEQUENCES AND SERIES 



Exercise 3-2 



1 . What is J2 2 ? 



2. Determine J2 *• 



3. Expand 2~Z *■ 

4. Calculate the value of a if: 



^a.2 fc_1 = 28 



f/VJ More practice CrJ video solutions f9 J or help at www.everythingmaths.co.za 



(1.)01cf (2.) 01 eg (3.)01ch (4.) Olci 




3.6 Finite Arithmetic Series 




Remember that an arithmetic sequence is a sequence of numbers, such that the difference between 
any term and the previous term is a constant number, d, called the constant difference: 



a„ = ai + d (n — 1) 



(3.18) 



where 



• n is the index of the sequence; 

• a„ is the n^-term of the sequence; 

• ai is the first term; 

• d is the common difference. 

When we sum a finite number of terms in an arithmetic sequence, we get a finite arithmetic series. 

A simple arithmetic sequence is when ai = 1 and d = in the general form (3.18); in other words all 
the terms in the sequence are 1: 

ai = ai + d (i — 1) 

= l + 0.(i-l) 

= 1 

{a,} = {1; 1; 1; 1; 1; ...} 

If we wish to sum this sequence from i = 1 to any positive integer n, we would write 

n n 

^ ai = ^l = l + l + H hi (n times) 



2S 



CHAPTER 3. SEQUENCES AND SERIES 



3.6 



Because all the terms are equal to 1, it means that if we sum to n we will be adding ?i-number of l's 
together, which is simply equal to n: 



Ei 



(3.19) 



Another simple arithmetic sequence is when <n = 1 and d = I, which is the sequence of positive 
integers: 

a; = ai + d (i — 1) 

= 1 + 1 . (* - 1) 

= i 

{a,} = {1; 2; 3; 4; 5; ...} 

If we wish to sum this sequence from i = 1 to any positive integer re, we would write 



Y^i= 1 + 2 + 3 



(3.20) 



This is an equation with a very important solution as it gives the answer to the sum of positive integers. 
We first write S„ as a sum of terms in ascending order: 



S„ = 1 + 2 + h (n - 1) + n 

We then write the same sum but with the terms in descending order: 

S n = n + {n - 1) + h 2 + 1 



(3.21) 



(3.22) 



We then add corresponding pairs of terms from Equations (3.21) and (3.22), and we find that the sum 
for each pair is the same, (n + 1): 

2 S n = (n + 1) + (n + 1) + • • • + (n + 1) + (n + 1) (3.23) 

We then have re-number of (n + l)-terms, and by simplifying we arrive at the final result: 

2 S n = re (re + 1) 
Sn = |(n + l) 



S n = ^2 i = - (n + 1) 



(3.24) 



Note that this is an example of a quadratic sequence. 



FACT 



Mathematician, Karl 

Friedrich Gauss, discov- 
ered this proof when 
he was only 8 years 
old. His teacher had 
decided to give his 
class a problem which 
would distract them 
for the entire day by 
asking them to add 
all the numbers from 
1 to 100. Young Karl 
realised how to do this 
almost instantaneously 
and shocked the teacher 
with the correct answer, 
5050. 




If we wish to sum any arithmetic sequence, there is no need to work it out term-for-term. We will now 
determine the general formula to evaluate a finite arithmetic series. We start with the general formula 



20 



3.6 



CHAPTER 3. SEQUENCES AND SERIES 



for an arithmetic sequence and sum it from i = 1 to any positive integer n: 

n n 

Y^a t = ^[oi+d(i-l)] 
i=l i=l 

n 

= / (a-i + di — d) 

n 

= Yl K ai - d )+ d{ \ 

n n 

= $^(ai-d) + X)(di) 

i=l i=l 

n n 

= J2 ( ai - d ) + d J2 l 

i a tin , , . 

= (ai — d)n+ — (n + 1) 

= - (2oi - 2d + dn + d) 

= — (2oi + dn — d) 

= ^[2oi+d(n-l)] 

So, the general formula for determining an arithmetic series is given by 

Sn = J2 [ai+d{i-l)] = ^ [2oi+d(n-l)] 



(3.25) 



For example, if we wish to know the series S20 for the arithmetic sequence a; = 3 + 7 (i — 1), we 
could either calculate each term individually and sum them: 



^'20 



£[3 + 7(i-l) 



= 3 + 10 + 17 + 24 + 31 + 38 + 45 + 52 + 
59 + 66 + 73 + 80 + 87 + 94 + 101 + 
108+ 115 + 122+ 129 + 136 

= 1390 

or, more sensibly, we could use Equation (3.25) noting that 01 = 3, d = 7 and n = 20 so that 

20 
S 20 = X> + 7(i-l)] 

= f [2.3 + 7(20-1)] 
= 1390 

This example demonstrates how useful Equation (3.25) is. 



Exercise 3-3 



1 . The sum to n terms of an arithmetic series is S n = — (In + 15). 

(a) How many terms of the series must be added to give a sum of 425? 

(b) Determine the (5 th term of the series. 

2. The sum of an arithmetic series is 100 times its first term, while the last term is 9 times the first 
term. Calculate the number of terms in the series if the first term is not equal to zero. 



30 



CHAPTER 3. SEQUENCES AND SERIES 



3.7 



3. The common difference of an arithmetic series is 3. Calculate the value of n for which the n th 
term of the series is 93, and the sum of the first n terms is 975. 

4. The sum of n terms of an arithmetic series is 5n 2 — lln for all values of n. Determine the 
common difference. 

5. The third term of an arithmetic sequence is —7 and the 7 th term is 9. Determine the sum of the 
first 51 terms of the sequence. 



6. Calculate the sum of the arithmetic series 4 + 7 + 10 + ■ ■ ■ + 901. 



A" 1 ) More practice CwJ video solutions Qfj or help at www.everythingmaths.co.za 



(1 .) Olcj (2.)01ck (3.) 01cm (4.) 01 en (5.) 01cp (6.) Olcq 




3.7 Finite Squared Series 




When we sum a finite number of terms in a quadratic sequence, we get a finite quadratic series. The 
general form of a quadratic series is quite complicated, so we will only look at the simple case when 
D = 2 and d = (02 — ai) = 3, where D is the common second difference and d is the first difference. 
This is the sequence of squares of the integers: 

■ 2 
o-i = 1 

{a,} = {l 2 ; 2 2 ; 3 2 ; 4 2 ; 5 2 ; 6 2 ; . . .} 

= {1; 4; 9; 16; 25; 36; ...} 

If we wish to sum this sequence and create a series, we write 

n 

Sn = J2 * 2 = 1 + 4 + 9 + ■■■ + n 

1 1 

which can be written, in general, as 



S„ = ^Y = ^(2n + l)(n + l) 
1=1 



The proof for Equation (3.26) can be found under the Extension block that follows: 



(3.26) 



Extension: 



Derivation of the Finite Squared Series 



We will now prove the formula for the finite squared series: 



S„ = ^y = l + 4 + 9 + --- + n 2 



We start off with the expansion of (k + 1)' 

(fc + 1) 3 



(jfc+l) 3 -fc 3 = 3/c 2 + 3fc + l 



31 



3.8 



CHAPTER 3. SEQUENCES AND SERIES 





k= 1 


: 2 3 


- l 3 = 3(1) 2 +3(1) + 1 




k = 2 : 3 3 


-2 3 = 3(2) 2 +3(2) + l 




k = 3 : 4 3 


-3 3 = 3(3) 2 +3(3) + l 




k = n : (n 


+ 1) 3 -n 3 = 3n 2 + 3n + l 




If we add all the terms on the right and left, we arrive at 




(„ + l) 3 - 1 = 


n 

J2 ( 3 * 2 + 3i + 1) 




n" + 3n~ + 'in + 1 — 1 = 


71 n n 

sE^E^ + E 1 

i=l i=l i=l 




n + 3n + 3n = 


n 

3^* 2 + T (n+l)+n 

1 = 1 




n 

E* 2 = 

I 1 


i [n 3 + 3n 2 + 3n - ^ (n + 1) - n] 




= 


1 /• 3 i q 2 , 3 2 3 . 

— (n + 3n + 3?i — — n — — n — n) 




= 


l f J,3 2,1 s 

— in H — n H — n.) 
3 V 2 2 ' 




= 


- (2n + 3n + 1) 
6 




Therefore, 










71 

E* 2 = 


£(2n + l)(n + l) 






















3.8 F/n/te Geometric Series 




When we sum a known number of terms in a geometric sequence, we get a finite geometric series. 
We can write out each term of a geometric sequence in the general form: 



a„ = ai . r 



(3.27) 



where 



n is the index of the sequence; 

a n is the n' h -term of the sequence; 

ai is the first term; 

r is the common ratio (the ratio of any term to the previous term). 



32 



CHAPTER 3. SEQUENCES AND SERIES 3.8 

By simply adding together the first n terms, we are actually writing out the series 

S n = ai + ai r + ai r + ■ ■ ■ + ai r n ~~ + ai r n ~ (3.28) 

We may multiply the above equation by r on both sides, giving us 

rSn = air + air + air + ■ ■ ■ + ai r n + Qi r" (3.29) 

You may notice that all the terms on the right side of (3.28) and (3.29) are the same, except the first 
and last terms. If we subtract (3.28) from (3.29), we are left with just 

rS„ — S„ = ai r n — ai 
S n {r-l) = ai(r n -l) 

Dividing by (r — 1) on both sides, we arrive at the general form of a geometric series: 

•A |_! ai (r n - 1) 

i = > ai.r = (3.30) 

i. — ( r — 1 

i=l 

See video: VMgjq at www.everythingmaths.co.za 



Exercise 3-4 



1 . Prove that 



-L ^2^ , n-1 a(l-r") 

a + ar + ar + ■ ■ • + ar = — — 

(1-r) 



2. Given the geometric sequence 1; —3; 9; . . . determine: 

(a) The 8 th term of the sequence 

(b) The sum of the first eight terms of the sequence. 

3. Determine: 

4 

,, i 

4. Find the sum of the first 11 terms of the geometric series 6 + 3+| + | + ... 

5. Show that the sum of the first n terms of the geometric series 

54+18 + 6 + --- + 5(i) n ~ 1 
is given by 81-3 4 "". 

6. The eighth term of a geometric sequence is 640. The third term is 20. Find the sum of the first 7 
terms. 

7. Solve for n; E8(|)' = 15§. 

4=1 

8. The ratio between the sum of the first three terms of a geometric series and the sum of the 4" 1 -, 
5"'- and 6"'-terms of the same series is 8 : 27. Determine the common ratio and the first 2 terms 
if the third term is 8. 



f/Vj More practice f ►) video solutions C"?) or help at www.everythingmaths.co.za 



(1.)01cr (2.) 01cs (3.)01ct (4.) 01cu (5.) 01cv (6.) 01cw 
(7.) 01 ex (8.)01cy 



33 



3.9 



CHAPTER 3. SEQUENCES AND SERIES 




3.9 Infinite Series 




Thus far we have been working only with finite sums, meaning that whenever we determined the 
sum of a series, we only considered the sum of the first n terms. In this section, we consider what 
happens when we add infinitely many terms together. You might think that this is a silly question - 
surely the answer will be oo when one sums infinitely many numbers, no matter how small they are? 
The surprising answer is that while in some cases one will reach oo (like when you try to add all the 
positive integers together), there are some cases in which one will get a finite answer. If you don't 
believe this, try doing the following sum, a geometric series, on your calculator or computer: 



You might think that if you keep adding more and more terms you will eventually get larger and larger 
numbers, but in fact you won't even get past 1 - try it and see for yourself! 

We denote the sum of an infinite number of terms of a sequence by 



Y. a > 



When we sum the terms of a series, and the answer we get after each summation gets closer and closer 
to some number, we say that the series converges. If a series does not converge, then we say that it 
diverges. 



Infinite Geometric Series 



EMCAC 



There is a simple test for knowing instantly which geometric series converges and which diverges. 
When r, the common ratio, is strictly between —1 and 1, i.e. — 1 < r < 1, the infinite series will 
converge, otherwise it will diverge. There is also a formula for working out the value to which the 
series converges. 

Let's start off with Formula (3.30) for the finite geometric series: 



I>i- 



,.«_! = oi (r" - 1) 
r-1 



Now we will investigate the behaviour of r" for — 1 < r < 1 as n becomes larger. 
Take r = = : 



n= 1 
n = 2 
n = 3 



y-2) 2 






(i) 2 = 2- 


1 

2 


4^2 


(k) 3 = k 


1 

2 


1 = i < 

2 8 ^ 



Since r is in the range — 1 < r < 1, we see that r" gets closer to as n gets larger. 



.31 



CHAPTER 3. SEQUENCES AND SERIES 



3.9 



Therefore, 



On 

'3 :^c 



ai(r" 


-1) 


r — 


1 


oi(0- 


-1) 


r — 


1 


— Ol 




r- 1 




ai 





for — 1 < r < 1 



1 -r 
The sum of an infinite geometric series is given by the formula 






for 



1 < r < 1 



where ax is the first term of the series and r is the common ratio. 
See video: VMgly at www.everythingmaths.co.za 



Exercise 3-5 



(3.31) 



1 . What does (§)" approach as n tends towards oo? 

2. Given the geometric series: 

2.(5) 5 +2.(5) 4 + 2.(5) ;i + ... 

(a) Show that the series converges 

(b) Calculate the sum to infinity of the series 

(c) Calculate the sum of the first eight terms of the series, correct to two decimal places. 

(d) Determine 

oo 

J2 2 ■ 5 6 ~" 
correct to two decimal places using previously calculated results. 

3. Find the sum to infinity of the geometric series 3+l + | + | + ... 

4. Determine for which values of x, the geometric series 

2 + f (x + 1) + | (x + if + . . . 
will converge. 

5. The sum to infinity of a geometric series with positive terms is 4| and the sum of the first two 
terms is 2|. Find a, the first term, and r, the common ratio between consecutive terms. 



A" 1 ) More practice CrJ video solutions f?J or help at www.everythingmaths.co.za 



(1.)01cz (2.) 01dO (3.) 01d1 (4.) 01d2 (5.) 01d3 



Chapter 3 



End of Chapter Exercises 



;)5 



3.9 CHAPTER 3. SEQUENCES AND SERIES 



1 . lsl + 2 + 3 + 4 + --- an example of a finite series or an infinite series? 

2. Calculate 



£3(i)* 



3. If x + 1; x — 1; 2x — 5 are the first three terms of a convergent geometric series, 
calculate the: 

(a) Value of x. 

(b) Sum to infinity of the series. 

4. Write the sum of the first twenty terms of the series 6 + 3+ § + § + ••• in ^-notation. 

5. For the geometric series, 

54+18 + 6 + --- + 5(|) n - 1 

calculate the smallest value of n for which the sum of the first n terms is greater than 
80.99. 

CO 

6. Determine the value of J2 12 (k) kl - 

k = l 

7. A new soccer competition requires each of 8 teams to play every other team once. 

(a) Calculate the total number of matches to be played in the competition. 

(b) If each of n teams played each other once, determine a formula for the total 
number of matches in terms of n. 

8. The midpoints of the opposite sides of square of length 4 units are joined to form 4 
new smaller squares. This midpoints of the new smaller squares are then joined in 
the same way to make even smaller squares. This process is repeated indefinitely. 
Calculate the sum of the areas of all the squares so formed. 

9. Thembi worked part-time to buy a Mathematics book which cost R29,50. On 1 
February she saved Rl,60, and everyday saves 30 cents more than she saved the 
previous day. (So, on the second day, she saved R1.90, and so on.) After how many 
days did she have enough money to buy the book? 

1 0. Consider the geometric series: 

5 + 2i + l| + ... 

(a) If A is the sum to infinity and B is the sum of the first n terms, write down the 
value of: 

i. A 
ii. B in terms of n. 

(b) For which values of n is (A — B) < ^? 

1 1 . A certain plant reaches a height of 118 mm after one year under ideal conditions in a 
greenhouse. During the next year, the height increases by 12 mm. In each successive 
year, the height increases by § of the previous year's growth. Show that the plant will 
never reach a height of more than 150 mm. 

n 

12. Calculate the value of n if J2 (20 - 4a) = -20. 

a=l 

13. Michael saved R400 during the first month of his working life. In each subsequent 
month, he saved 10% more than what he had saved in the previous month. 

(a) How much did he save in the 7 th working month? 

(b) How much did he save all together in his first 12 working months? 

(c) In which month of his working life did he save more than Rl 500 for the first 
time? 

14. A man was injured in an accident at work. He receives a disability grant of R4 800 in 
the first year. This grant increases with a fixed amount each year. 



36 



CHAPTER 3. SEQUENCES AND SERIES 3.9 



(a) What is the annual increase if, over 20 years, he would have received a total of 
R143 500? 

(b) His initial annual expenditure is R2 600 and increases at a rate of R400 per year. 
After how many years does his expenses exceed his income? 

15. The Cape Town High School wants to build a school hall and is busy with fundraising. 
Mr. Manuel, an ex-learner of the school and a successful politician, offers to donate 
money to the school. Having enjoyed mathematics at school, he decides to donate 
an amount of money on the following basis. He sets a mathematical quiz with 20 
questions. For the correct answer to the first question (any learner may answer), the 
school will receive 1 cent, for a correct answer to the second question, the school 
will receive 2 cents, and so on. The donations 1; 2; 4: . . . form a geometric sequence. 
Calculate (Give your answer to the nearest Rand) 

(a) The amount of money that the school will receive for the correct answer to the 
20 th question. 

(b) The total amount of money that the school will receive if all 20 questions are 
answered correctly. 

16. The common difference of an arithmetic series is 3. Calculate the values of n for 
which the n th term of the series is 93 and the sum of the first n terms is 975. 

1 7. The first term of a geometric sequence is 9, and the ratio of the sum of the first eight 
terms to the sum of the first four terms is 97 : 81. Find the first three terms of the 
sequence, if it is given that all the terms are positive. 

18. (fe — 4); (fc + 1); m; 5fc is a set of numbers, the first three of which form an arithmetic 
sequence, and the last three a geometric sequence. Find k and m if both are positive. 

19. Given: The sequence 6 + p; 10 + p: 15 + p is geometric. 

(a) Determine p. 

(b) Show that the common ratio is f . 

(c) Determine the 10"* term of this sequence correct to one decimal place. 

20. The second and fourth terms of a convergent geometric series are 36 and 16, respec- 
tively. Find the sum to infinity of this series, if all its terms are positive. 

n , c . , 5 k(k + l) 

21. Eva uate: V — 

fc = 2 z 

22. S„ = An 1 + 1 represents the sum of the first n terms of a particular series. Find the 
second term. 

oo 12 

23. Find p if: £ 27p fc = £ (24-3*) 

fc=i t=i 

24. Find the integer that is the closest approximation to: 



102002 _|_ 1Q2002 



25. In each case (substituting the values of x below), determine if the series J2( x + 2) p 

P =i 
converges. If it does, work out what it converges to: 

(a)x = -5 

(b) x = -5 

26. Calculate: £ 5.4^ 

27. The sum of the first p terms of a sequence is p (p + 1). Find the 10 th term. 

28. The powers of 2 are removed from the set of positive integers 

1; 2; 3; 4; 5; 6; ... ; 1998; 1999; 2000 
Find the sum of remaining integers. 



37 



3.9 



CHAPTER 3. SEQUENCES AND SERIES 



29. Observe the pattern below: 




(a) If the pattern continues, find the number of letters in the column containing M's. 

(b) If the total number of letters in the pattern is 361, which letter will be in the last 
column. 

30. The following question was asked in a test: 

Find the value of 2 2005 + 2 2005 . 

Here are some of the students' answers: 



31 



(a) Megan said the answer is 4 2005 . 

(b) Stefan wrote down 2 4010 . 

(c) Nina thinks itis2 2006 . 

(d) Annatte gave the answer 2 2005x2005 . 

Who is correct? ("None of them" is also a possibility.) 

A shrub of height 110 cm is planted. At the end of the first year, the shrub is 120 cm 
tall. Thereafter, the growth of the shrub each year is half of its growth in the previous 
year. Show that the height of the shrub will never exceed 130 cm. 



More practice (►) video solutions ({J or help at www.everythingmaths.co.za 



(1.) 01d4 
(7.) 01 da 
(13.)01dg 
(19.) 01dn 
(25.) 01du 
(31.) 01e0 



(2.) 01 d5 
(8.) 01 db 
(14.) 01dh 
(20.) 01 dp 
(26.) 01 dv 



(3.) 01d6 
(9.) 01 dc 
(15.) 01di 
(21.) 01dq 
(27.) 01 dw 



(4.) 01d7 
(10.) 01dd 
(16.) 01dj 
(22.) 01dr 
(28.) 01 dx 



(5.) 01d8 
(11.) 01de 
(17.) 01dk 
(23.) 01ds 
(29.) 01 dy 



(6.) 01 d9 
(12.) 01df 
(18.) 01dm 
(24.) 01dt 
(30.) 01dz 



:is 



Finance 





4. 1 Introduction 




In earlier grades simple interest and compound interest were studied, together with the concept of 
depreciation. Nominal and effective interest rates were also described. Since this chapter expands on 
earlier work, it would be best if you revised the work done in Grades 10 and 1 1, 

When you master the techniques in this chapter, you will be able to assess critically how to invest 
your money when you start working and earning. And when you are looking at applying for a bond 
from a bank to buy a home, you will confidently be able to get out the calculator and work out how 
much you could actually save by making additional repayments. This chapter will provide you with 
the fundamental concepts you will need to manage your finances. 

© See introductory video: VMgmf at www.everythingmaths.co.za 




4.2 Finding the Length of the 
Investment or Loan 



In Grade 1 1, we used the Compound Interest formula A = P(l + i) n to determine the term of the 
investment or loan, by trial and error. Remember that P is the initial amount, A is the current amount, 
i is the interest rate and n is the number of time units (number of months or years). So if we invest an 
amount and know what the interest rate is, then we can work out how long it will take for the money 
to grow to the required amount. 

Now that you have learnt about logarithms, you are ready to work out the proper algebraic solution. If 
you need to remind yourself how logarithms work, go to Chapter 2 (on Page 4). 

The basic finance equation is: 

A = P.(l + i) n 

If you don't know what A, P, i and n represent, then you should definitely revise the work from Grade 
10 and 11. 

Solving for n: 

A = P{l + i) n 

(l + i)» = (A/P) 

log((l+*) n ) = log(A/P) 

nlog(l + i) = log(A/P) 

n = log(A/P)/log(l + i) 

Remember, you do not have to memorise this formula. It is very easy to derive any time you need 
it. It is simply a matter of writing down what you have, deciding what you need, and solving for that 
variable. 



39 



4.3 



CHAPTER 4. FINANCE 



Example 1: Term of Investment - Logarithms 



QUESTION 


Suppose we invested R3 500 into a savings account which pays 7,5% compound interest. After 
an unknown period of time our account is worth R4 044,69. For how long did we invest the 
money? How does this compare with the trial and error answer from Chapter ??. 


SOLUTION 






Step 7 : 


Determine what is given and what is required 

• P = R3 500 






• i = 7,5% 








. A = R4 044,69 
We are required to find n 






Step 2 : 


Determine how to approach the problem 

We know that: 






A = 


P(l + i) n 






(1 + 0" = 


(A/P) 






bg((l + i)") = 


log(A/P) 






nlog(l + i) = 


log(A/P) 






n = 


log(A/P)/log(l + i) 




Step 3 : 


Solve the problem 








n = log(A/P)/log(l + ») 

logf 4044 ' 69 ) 7 5 
= , , 3 ™ ' Remember that: 7.5% ='„„= 0.075 
log(l + 0.075) 100 

= 2.0 




Step 4 : 


Write final answer 

The R3 500 was invested for 2 


years. 








4.3 Series of Payments 




By this stage, you know how to do calculations such as 'If I want Rl 000 in three years' time, how 
much do I need to invest now at 10%?' 

What if we extend this as follows: "If I want to draw Rl 000 next year, Rl 000 the next year and Rl 000 



10 



CHAPTER 4. FINANCE 4.3 



after three years, how much do I need to initially put into a bank account earning 10% p.a. to be able 
to afford to be able to do this?" 

The obvious way of working that out is to work out how much you need now to afford the pay- 
ments individually and sum them. We'll work out how much is needed now to afford the payment of 
Rl 000 in a year (= Rl 000 x (1,10) _1 =R909,09), the amount needed now for the following year's 
Rl 000(=R1 000 x (1,10)~ 2 = R826,45) and the amount needed now for the Rl 000 after three years 
(=R1 000 x (1,10)~ 3 =R751,31). Adding these together gives you the amount needed to afford all 
three payments and you get R2 486,85. 

So, if you put R2 486,85 into a 10% bank account now, you will be able to draw out Rl 000 in a year, 
Rl 000 a year after that, and Rl 000 a year after that - and your bank account balance will decrease to 
R0. You would have had exactly the right amount of money to do that (obviously!). 

You can check this as follows: 

Amount at Time (i.e. Now) = R2 486,85 

Amount at Time 1 (i.e. a year later) = R2 486,85(1 + 10%) = R2 735,54 

Amount after withdrawing Rl 000 = R2 735,54 - Rl 000 = Rl 735,54 

Amount at Time 2 (i.e. a year later) = Rl 735,54(1 + 10%) = Rl 909,09 

Amount after withdrawing Rl 000 = Rl 909,09 - Rl 000 = R909,09 

Amount at Time 3 (i.e. a year later) = R909,09(l + 10%) = Rl 000 

Amount after withdrawing Rl 000 = Rl 000 - Rl 000 = R0 

Perfect! Of course, for only three years, that was not too bad. But what if I asked you how much you 
needed to put into a bank account now, to be able to afford R100 a month for the next 1 5 years. If you 
used the above approach you would still get the right answer, but it would take you weeks! 

There is - I'm sure you guessed - an easier way! This section will focus on describing how to work 
with: 

• annuities - a fixed sum payable each year or each month, either to provide a pre-determined 
sum at the end of a number of years or months (referred to as a future value annuity) or a fixed 
amount paid each year or each month to repay (amortise) a loan (referred to as a present value 
annuity). 

• bond repayments - a fixed sum payable at regular intervals to pay off a loan. This is an example 
of a present value annuity. 

• sinking funds - an accounting term for cash set aside for a particular purpose and invested so that 
the correct amount of money will be available when it is needed. This is an example of a future 
value annuity. 



Sequences and Series W emcag 



Before we progress, you need to go back and read Chapter 3 (from Page 19) to revise sequences and 
series. 

In summary, if you have a series of n terms in total which looks like this: 

a + ar + ar + ■ ■ ■ + ar = a[l + r + r + ■ ■ ■ r 

this can be simplified as: 

a{r n - 1) , , , 

— useful when r > 1 

r — 1 

all — r") 

— useful when < r < 1 

1 — r 



11 



4.3 CHAPTER 4. FINANCE 



Present Values of a Series of Payments W emcah 



So having reviewed the mathematics of sequences and series, you might be wondering how this is 
meant to have any practical purpose! Given that we are in the finance section, you would be right to 
guess that there must be some financial use to all this. Here is an example which happens in many 
people's lives - so you know you are learning something practical. 

Let us say you would like to buy a property for R300 000, so you go to the bank to apply for a mortgage 
bond. The bank wants it to be repaid by annually payments for the next 20 years, starting at end of 
this year. They will charge you 15% interest per annum. At the end of the 20 years the bank would 
have received back the total amount you borrowed together with all the interest they have earned from 
lending you the money. You would obviously want to work out what the annual repayment is going to 
be! 

Let X be the annual repayment, i is the interest rate, and M is the amount of the mortgage bond you 
will be taking out. 

Time lines are particularly useful tools for visualising the series of payments for calculations, and we 
can represent these payments on a time line as: 



XX XXX 

t t i ■ ' iii T c : h e Flows 

12 18 19 20 



Figure 4.1: Time line for an annuity (in arrears) of X for n periods. 

The present value of all the payments (which includes interest) must equate to the (present) value of 
the mortgage loan amount. 

Mathematically, you can write this as: 

M = X(l + i)- 1 + X(l + €)- 2 + X{1 + i)- 3 + ■ ■ ■ + X(l + i)- 20 

The painful way of solving this problem would be to do the calculation for each of the terms above - 
which is 20 different calculations. Not only would you probably get bored along the way, but you are 
also likely to make a mistake. 

Naturally, there is a simpler way of doing this! You can rewrite the above equation as follows: 



M = Xlv 1 + v 2 + v 3 + h v 20 } 

where v = (1 + «) _1 = 1/(1 + i) 

Of course, you do not have to use the method of substitution to solve this. We just find this a useful 
method because you can get rid of the negative exponents - which can be quite confusing! As an 
exercise - to show you are a real financial whizz - try to solve this without substitution. It is actually 
quite easy. 

Now, the item in square brackets is the sum of a geometric sequence, as discussion in section 3. This 



12 



CHAPTER 4. FINANCE 



4.3 



can be re-written as follows, using what we know from Chapter 3 of this text book: 

v(l + v + v + • ■ ■ + v n ~ ) 
' \-v n S 
1 ~ 
l-v n 
l/v- 1 

l-(i + Q- 

i 



Note that we took out a common factor of v before using the formula for the geometric sequence. 
So we can write: 

M = X 



(l-(l + i)-») 



This can be re-written: 



X 



M 



Mi 



(i-(i + Q-")l 



(l + i)- n 



So, this formula is useful if you know the amount of the mortgage bond you need and want to work 
out the repayment, or if you know how big a repayment you can afford and want to see what property 
you can buy. 

For example, if I want to buy a house for R300 000 over 20 years, and the bank is going to charge me 
15% per annum on the outstanding balance, then the annual repayment is: 



.Y 



Mi 



1- (1 + i)"" 
R300 000x0,15 

1-(1 + 0,15)- 20 
R47 928.44 



This means, each year for the next 20 years, I need to pay the bank R47 928,44 per year before I have 
paid off the mortgage bond. 

On the other hand, if I know I will have only R30 000 per year to repay my bond, then how big a house 
can I buy? That is easy. 



M 



X 



(i-(i + 0-) 



R30 000 



(1 - (1,15)- 20 ) 



0,15 



R187 779,94 



So, for R30 000 a year for 20 years, I can afford to buy a house of R187 800 (rounded to the nearest 
hundred). 

The bad news is that R187 800 does not come close to the R300 000 you wanted to pay! The good 
news is that you do not have to memorise this formula. In fact , when you answer questions like this in 
an exam, you will be expected to start from the beginning - writing out the opening equation in full, 
showing that it is the sum of a geometric sequence, deriving the answer, and then coming up with the 
correct numerical answer. 



1.3 



4.3 CHAPTER 4. FINANCE 



Example 2: Monthly mortgage repayments 



QUESTION 



Sam is looking to buy his first flat, and has R15 000 in cash savings which he will use as a 
deposit. He has viewed a flat which is on the market for R250 000, and he would like to work 
out how much the monthly repayments would be. He will be taking out a 30 year mortgage 
with monthly repayments. The annual interest rate is 11%. 



SOLUTION 



Step 1 : Determine what is given and what is needed 

• Deposit amount = R15 000 

The following is given: • Price of flat = R250 000 

• interest rate, i = 11% 

We are required to find the monthly repayment for a 30-year mortgage. 

Step 2 : Determine how to approach the problem 

We know that: 

X: M 



(l-(l + i)-") l 



In order to use this equation, we need to calculate M, the amount of the mortgage 
bond, which is the purchase price of property less the deposit which Sam pays 
upfront. 

M = R250 000 - R15 000 
= R235 000 

Now because we are considering monthly repayments, but we have been 
given an annual interest rate, we need to convert this to a monthly interest rate, 
ji2. (If you are not clear on this, go back and revise Section ??.) 



(i + ii 2 r 


= (i + ») 


(1 + J12) 12 


= 1,11 


ii2 


= 0,873459% 



We know that the mortgage bond is for 30 years, which equates to 360 
months. 

Step 3 : Solve the problem 

Now it is easy, we can just plug the numbers in the formula, but do not forget 
that you can always deduce the formula from first principles as well! 



X 



M 



r (l-(l + i)-") i 

R235 000 

r (l-(1.00876459)- 360 ) i 
I 0,008734594 J 

R2 146,39 



Step 4 : Write the final answer 



li 



CHAPTER 4. FINANCE 4.3 



That means that to buy a flat for R250 000, after Sam pays a R15 000 deposit, 
he will make repayments to the bank each month for the next 30 years equal to 
R2 146,39. 



Example 3: Monthly mortgage repayments 



QUESTION 



You are considering purchasing a flat for R200 000 and the bank's mortgage rate is currently 
9% per annum payable monthly. You have savings of R10 000 which you intend to use for 
a deposit. How much would your monthly mortgage payment be if you were considering a 
mortgage over 20 years. 



SOLUTION 



Step 1 : Determine what is given and what is required 

The following is given: 

• Deposit amount = R10 000 

• Price of flat = R200 000 

• Interest rate, i = 9% 

We are required to find the monthly repayment for a 20-year mortgage. 

Step 2 : Determine how to approach the problem 

We are considering monthly mortgage repayments, so it makes sense to use 
months as our time period. 

The interest rate was quoted as 9% per annum payable monthly, which 
means that the monthly effective rate = 5J = 0,75% per month. Once we 
have converted 20 years into 240 months, we are ready to do the calculations! 

First we need to calculate M, the amount of the mortgage bond, which is 
the purchase price of property minus the deposit which Sam pays up-front. 

M = R200 000 - R10 000 
= R190 000 

The present value of our mortgage payments X (which includes interest), 
must equate to the present mortgage amount 

M = X x (H-0,75%) -1 + 

X x (l + 0,75%r 2 + 

X x (l + 0,75%r 3 + 

X x (l + 0,75%r 4 + ... 
X x (1 + 0,75%r 239 + X x (1 + 0,75%r 240 

But it is clearly much easier to use our formula than work out 240 factors 
and add them all up! 

Step 3 : Solve the problem 



15 



4.3 CHAPTER 4. FINANCE 



X x 111,14495 = R190 000 
X = Rl 709,48 



Step 4 : Write f/ie ffna/ answer 

So to repay a R190 000 mortgage over 20 years, at 9% interest payable monthly, 
will cost you Rl 709,48 per month for 240 months. 



Show Me the Money 

Now that you've done the calculations for the worked example and know what the monthly repayments 
are, you can work out some surprising figures. For example, Rl 709,48 per month for 240 months 
makes for a total of R410 275,20 (=R1 709,48 x 240). That is more than double the amount that you 
borrowed! This seems like a lot. However, now that you've studied the effects of time (and interest) on 
money, you should know that this amount is somewhat meaningless. The value of money is dependant 
on its timing. 

Nonetheless, you might not be particularly happy to sit back for 20 years making your Rl 709,48 
mortgage payment every month knowing that half the money you are paying are going toward interest. 
But there is a way to avoid those heavy interest charges. It can be done for less than R300 extra every 
month. 

So our payment is now R2 000. The interest rate is still 9% per annum payable monthly (0,75% per 
month), and our principal amount borrowed is R190 000. Making this higher repayment amount every 
month, how long will it take to pay off the mortgage? 

The present value of the stream of payments must be equal to R190 000 (the present value of the 
borrowed amount). So we need to solve for n in: 



R2 000 x [1 - (1 + 0,75%r n ]/0,75% = R190 000 

1- (1 + 0,75%)- = ( 19000 2 X 00 °' 75% ) 

lorfl + 0,75%)- = log[(l- 19000 2 ° X 00 ' 0075 ] 

, ,-, „, r »,> , r/-, 190 000x0,0075, 

-nxlog(l + 0,75%) = log[(l ^^ ] 

-n x 0,007472 = -1,2465 

n = 166,8 months 

= 13,9 years 



So the mortgage will be completely repaid in less than 14 years, and you would have made a total 
payment of 166,8 x R2 000 = R333 600. 

Can you see what is happened? Making regular payments of R2 000 instead of the required Rl 709,48, 
you will have saved R76 675,20 (= R410 275,20- R333 600) in interest, and yet you have only paid 
an additional amount of R290,52 for 1 66,8 months, or R48 458,74. You surely know by now that the 
difference between the additional R48 458,74 that you have paid and the R76 675,20 interest that you 
have saved is attributable to, yes, you have got it, compound interest! 



10 



CHAPTER 4. FINANCE 4.3 



Future Value of a Series of Payments mEMCAi 



In the same way that when we have a single payment, we can calculate a present value or a future 
value - we can also do that when we have a series of payments. 

In the above section, we had a few payments, and we wanted to know what they are worth now - 
so we calculated present values. But the other possible situation is that we want to look at the future 
value of a series of payments. 

Maybe you want to save up for a car, which will cost R45 000 - and you would like to buy it in 2 years 
time. You have a savings account which pays interest of 12% per annum. You need to work out how 
much to put into your bank account now, and then again each month for 2 years, until you are ready 
to buy the car. 

Can you see the difference between this example and the ones at the start of the chapter where we 
were only making a single payment into the bank account - whereas now we are making a series of 
payments into the same account? This is a sinking fund. 

So, using our usual notation, let us write out the answer. Make sure you agree how we come up with 
this. Because we are making monthly payments, everything needs to be in months. So let A be the 
closing balance you need to buy a car, P is how much you need to pay into the bank account each 
month, and i r2 is the monthly interest rate. (Careful - because 12% is the annual interest rate, so we 
will need to work out later what the monthly interest rate is!) 

A = P(1 + in) 24 + P(l + «i2) 23 + ■ ■ ■ + P(l + J12) 1 
Here are some important points to remember when deriving this formula: 

1. We are calculating future values, so in this example we use (1 + ir 2 ) n and not (1 + 112)"". 
Check back to the start of the chapter if this is not obvious to you by now. 

2. If you draw a timeline you will see that the time between the first payment and when you buy 
the car is 24 months, which is why we use 24 in the first exponent. 

3. Again, looking at the timeline, you can see that the 24" 1 payment is being made one month 
before you buy the car - which is why the last exponent is a 1. 

4. Always check that you have got the right number of payments in the equation. Check right now 
that you agree that there are 24 terms in the formula above. 

So, now that we have the right starting point, let us simplify this equation: 

A = P[(l + j 12 ) 24 + (l + i 12 ) 23 + ... + (l-H 12 ) 1 ] 

= P[X 2A + X 23 + ■■■ + X 1 ] using X = (1 + i 12 ) 

Note that this time X has a positive exponent not a negative exponent, because we are doing future 
values. This is not a rule you have to memorise - you can see from the equation what the obvious 
choice of X should be. 

Let us re-order the terms: 

A = PIX 1 + X 2 H h X 24 ] = P . X[l + X + X 2 + h X 23 ] 

This is just another sum of a geometric sequence, which as you know can be simplified as: 

A = P.X[X n -l]/((l + i 12 )-l) 
= P.X{X n -l]/i 12 



4.4 



CHAPTER 4. FINANCE 



So if we want to use our numbers, we know that A =R45 000, n = 24 (because we are looking at 
monthly payments, so there are 24 months involved) and i = 12% per annum. 

BUT (and it is a big but) we need a monthly interest rate. Do not forget that the trick is to keep the time 
periods and the interest rates in the same units - so if we have monthly payments, make sure you use a 
monthly interest rate! Using the formula from Grade 1 1, we know that (1 +i) = (1 4- J12) 12 . So we 
can show that i 12 = 0,0094888 = 0,94888%. 

Therefore, 

45 000 = P(l,0094888)[(l,0094888) 24 - l]/0,0094888 
P = 1662,67 



This means you need to invest R166 267 each month into that bank account to be able to pay for your 
car in 2 years time. 



Exercise 4-1 



1 . You have taken out a mortgage bond for R875 000 to buy a flat. The bond is for 30 years and the 
interest rate is 12% per annum payable monthly. 

(a) What is the monthly repayment on the bond? 

(b) How much interest will be paid in total over the 30 years? 

2. How much money must be invested now to obtain regular annuity payments of R5 500 per 
month for five years? The money is invested at 11,1% p. a., compounded monthly. (Answer to 
the nearest hundred rand.) 



A"j More practice f ►) video solutions Cf) or help at www.everythingmaths.co.za 



(1.)01e1 (2.) 01e2 




4.4 Investments and Loans 




By now, you should be well equipped to perform calculations with compound interest. This section 
aims to allow you to use these valuable skills to critically analyse investment and loan options that you 
will come across in your later life. This way, you will be able to make informed decisions on options 
presented to you. 

At this stage, you should understand the mathematical theory behind compound interest. However, 
the numerical implications of compound interest are often subtle and far from obvious. 

Recall the example 'Show Me the Money' in Section 4.3.2. For an extra payment of R29 052 a month, 
we could have paid off our loan in less than 14 years instead of 20 years. This provides a good 
illustration of the long term effect of compound interest that is often surprising. In the following 
section, we'll aim to explain the reason for the drastic reduction in the time it takes to repay the loan. 



IS 



CHAPTER 4. FINANCE 4.4 



Loan Schedules ■ emcak 



So far, we have been working out loan repayment amounts by taking all the payments and discounting 
them back to the present time. We are not considering the repayments individually. Think about the 
time you make a repayment to the bank. There are numerous questions that could be raised: how 
much do you still owe them? Since you are paying off the loan, surely you must owe them less money, 
but how much less? We know that we'll be paying interest on the money we still owe the bank. When 
exactly do we pay interest? How much interest are we paying? 

The answer to these questions lie in something called the load schedule. 

We will continue to use the earlier example. There is a loan amount of R190 000. We are paying it 
off over 20 years at an interest of 9% per annum payable monthly. We worked out that the repayments 
should be Rl 709,48. 

Consider the first payment of Rl 709,48 one month into the loan. First, we can work out how much 
interest we owe the bank at this moment. We borrowed R190 000 a month ago, so we should owe: 



M x iia 

R190 000 x 0,75% 

R1425 



We are paying them Rl 425 in interest. We call this the interest component of the repayment. We are 
only paying off Rl 709,48— Rl 425 =R284,48 of what we owe! This is called the capital component. 
That means we still owe R190 000- R284,48 = R189 715,52. This is called the capital outstanding. 
Let's see what happens at the end of the second month. The amount of interest we need to pay is the 
interest on the capital outstanding. 



M x ii2 

R189 715,52 x 0,75% 

Rl 422,87 



Since we don't owe the bank as much as we did last time, we also owe a little less interest. The capital 
component of the repayment is now Rl 709,48— Rl 422,87 = R286,61. The capital outstanding will 
be R189 715,52- R286,61 = R189 428,91. This way, we can break each of our repayments down into 
an interest part and the part that goes towards paying off the loan. 

This is a simple and repetitive process. Table 4.1 is a table showing the breakdown of the first 12 
payments. This is called a loan schedule. 

Now, let's see the same thing again, but with R2 000 being repaid each year. We expect the numbers 
to change. However, how much will they change by? As before, we owe Rl 425 in interest in interest. 
After one month. However, we are paying R2 000 this time. That leaves R575 that goes towards paying 
off the capital outstanding, reducing it to R189 425. By the end of the second month, the interest owed 
is Rl 420,69 (That's R189 425 x i 12 ). Our R2 000 pays for that interest, and reduces the capital amount 
owed by R2 000- Rl 420,69 = R579,31. This reduces the amount outstanding to R188 845,69. 

Doing the same calculations as before yields a new loan schedule shown in Table 4.2. 

The important numbers to notice is the "Capital Component" column. Note that when we are paying 
off R2 000 a month as compared to Rl 709,48 a month, this column more than double. In the beginning 
of paying off a loan, very little of our money is used to pay off the capital outstanding. Therefore, even 



19 



4.4 



CHAPTER 4. FINANCE 



Time 


Repayment 


Interest Com- 


Capital Com- 


Capital Outstand- 








ponent 


ponent 


ing 



















R 


190 000,00 


1 


R 


1 709,48 


R 


1 425,00 


R 


284,48 


R 


189 715,52 


2 


R 


1 709,48 


R 


1 422,87 


R 


286,61 


R 


189 428,91 


3 


R 


1 709,48 


R 


1 420,72 


R 


288,76 


R 


189 140,14 


4 


R 


1 709,48 


R 


1 418,55 


R 


290,93 


R 


188 849,21 


5 


R 


1 709,48 


R 


1 416,37 


R 


293,11 


R 


188 556,10 


6 


R 


1 709,48 


R 


1414,17 


R 


295,31 


R 


188 260,79 


7 


R 


1 709,48 


R 


1411,96 


R 


297,52 


R 


187 963,27 


8 


R 


1 709,48 


R 


1 409,72 


R 


299,76 


R 


187 663,51 


9 


R 


1 709,48 


R 


1 407,48 


R 


302,00 


R 


187 361,51 


10 


R 


1 709,48 


R 


1 405,21 


R 


304,27 


R 


187 057,24 


11 


R 


1 709,48 


R 


1 402,93 


R 


306,55 


R 


186 750,69 


12 


R 


1 709,48 


R 


1 400,63 


R 


308,85 


R 


186 441,84 



Table 4.1 : A loan schedule with repayments of Rl 709,48 per month. 



Time 


Repayment 


Interest Com- 


Capital Com- 


Capital Outstand- 








ponent 


ponent 


ing 



















R 


190 000,00 


1 


R 


2 000,00 


R 


1 425,00 


R 


575,00 


R 


189 425,00 


2 


R 


2 000,00 


R 


1 420,69 


R 


579,31 


R 


188 845,69 


3 


R 


2 000,00 


R 


1 416,34 


R 


583,66 


R 


188 262,03 


4 


R 


2 000,00 


R 


1411,97 


R 


588,03 


R 


187 674,00 


5 


R 


2 000,00 


R 


1 407,55 


R 


592,45 


R 


187 081,55 


6 


R 


2 000,00 


R 


1 403,11 


R 


596,89 


R 


186 484,66 


7 


R 


2 000,00 


R 


1 398,63 


R 


601,37 


R 


185 883,30 


8 


R 


2 000,00 


R 


1 394,12 


R 


605,88 


R 


185 277,42 


9 


R 


2 000,00 


R 


1 389,58 


R 


610,42 


R 


184 667,00 


10 


R 


2 000,00 


R 


1 385,00 


R 


615,00 


R 


184 052,00 


11 


R 


2 000,00 


R 


1 380,39 


R 


619,61 


R 


183 432,39 


12 


R 


2 000,00 


R 


1 375,74 


R 


624,26 


R 


182 808,14 



Table 4.2: A loan schedule with repayments of R2 000 per month. 



50 



CHAPTER 4. FINANCE 4.4 



a small increase in repayment amounts can significantly increase the speed at which we are paying off 
the capital. 

What's more, look at the amount we are still owing after one year (i.e. at time 12). When we were 
paying Rl 709,48 a month, we still owe R186 441,84. However, if we increase the repayments to 
R2 000 a month, the amount outstanding decreases by over R3 000 to R182 808,14. This means we 
would have paid off over R7 000 in our first year instead of less than R4 000. This increased speed at 
which we are paying off the capital portion of the loan is what allows us to pay off the whole loan 
in around 14 years instead of the original 20. Note however, the effect of paying R2 000 instead of 
Rl 709,48 is more significant in the beginning of the loan than near the end of the loan. 

It is noted that in this instance, by paying slightly more than what the bank would ask you to pay, you 
can pay off a loan a lot quicker. The natural question to ask here is: why are banks asking us to pay the 
lower amount for much longer then? Are they trying to cheat us out of our money? 

There is no simple answer to this. Banks provide a service to us in return for a fee, so they are out to 
make a profit. However, they need to be careful not to cheat their customers for fear that they'll simply 
use another bank. The central issue here is one of scale. For us, the changes involved appear big. We 
are paying off our loan 6 years earlier by paying just a bit more a month. To a bank, however, it doesn't 
matter much either way. In all likelihood, it doesn't affect their profit margins one bit! 

Remember that the bank calculates repayment amounts using the same methods as we've been learn- 
ing. They decide on the correct repayment amounts for a given interest rate and set of terms. Smaller 
repayment amounts will make the bank more money, because it will take you longer to pay off the 
loan and more interest will accumulate. Larger repayment amounts mean that you will pay off the loan 
faster, so you will accumulate less interest i.e. the bank will make less money off of you. It's a simple 
matter of less money now or more money later. Banks generally use a 20 year repayment period by 
default. 

Learning about financial mathematics enables you to duplicate these calculations for yourself. This 
way, you can decide what's best for you. You can decide how much you want to repay each month 
and you'll know of its effects. A bank wouldn't care much either way, so you should pick something 
that suits you. 



Example 4: Monthly Payments 



QUESTION 



Stefan and Mama want to buy a house that costs R 1 200 000. Their parents offer to put down 
a 20% payment towards the cost of the house. They need to get a mortgage for the balance. 
What are their monthly repayments if the term of the home loan is 30 years and the interest is 
7,5%, compounded monthly? 



SOLUTION 



Step 1 : Determine how much money they need to borrow 

Rl 200 00-R240 000 =R960 000 

Step 2 : Determine how to approach the problem 

Use the formula: 

p _ x[l-(l + i)-"] 



Where 

P =R960 000 



4.4 



CHAPTER 4. FINANCE 



n = 30 x 12 = 360 months 
4 = 0,075-^12 = 0,00625 



Step 3 : Solve the problem 



R960 000 



x[l-(l + 0,00625)~ 360 ] 

0,00625 
x(143,0176273) 
R6 712,46 



Step 4 : Write the final answer 

The monthly repayments =R6 712,46 



Exercise 4-2 



1. A property costs Rl 800 000. Calculate the monthly repayments if the interest rate is 14% p. a. 
compounded monthly and the loan must be paid off in 20 years time. 

2. A loan of R 4 200 is to be returned in two equal annual instalments. If the rate of interest of 10% 
per annum, compounded annually, calculate the amount of each instalment. 



f/Vj More practice Crj video solutions C ?J or help at www.everythingmaths.co.za 



(1.)01e3 (2.)01e4 



Calculating Capital Outstanding 



EMCAL 



As defined in Section 4.4.1, capital outstanding is the amount we still owe the people we borrowed 
money from at a given moment in time. We also saw how we can calculate this using loan schedules. 
However, there is a significant disadvantage to this method: it is very time consuming. For example, in 
order to calculate how much capital is still outstanding at time 12 using the loan schedule, we'll have 
to first calculate how much capital is outstanding at time 1 through to 11 as well. This is already quite 
a bit more work than we'd like to do. Can you imagine calculating the amount outstanding after 10 
years (time 120)? 

Fortunately, there is an easier method. However, it is not immediately clear why this works, so let's 
take some time to examine the concept. 



Prospective Method for Capital Outstanding 

Let's say that after a certain number of years, just after we made a repayment, we still owe amount Y. 
What do we know about Yl We know that using the loan schedule, we can calculate what it equals 
to, but that is a lot of repetitive work. We also know that Y is the amount that we are still going to pay 



CHAPTER 4. FINANCE 



4.5 



off. In other words, all the repayments we are still going to make in the future will exactly pay off Y. 
This is true because in the end, after all the repayments, we won't be owing anything. 

Therefore, the present value of all outstanding future payments equal the present amount outstanding. 
This is the prospective method for calculating capital outstanding. 

Let's return to a previous example. Recall the case where we were trying to repay a loan of R200 000 
over 20 years. A RIO 000 deposit was put down, so the amount being payed off was R190 000. At an 
interest rate of 9% compounded monthly, the monthly repayment was Rl 709,48. In Table 4.1, we can 
see that after 12 months, the amount outstanding was R186 441,84. Let's try to work this out using the 
the prospective method. 

After time 12, there are still 19 x 12 = 228 repayments left of Rl 709,48 each. The present value is: 



228 
0,75% 



Y 



Rl 709,48 x 
R186 441,92 



1 - 1,0075' 
0,0075 



Oops! This seems to be almost right, but not quite. We should have got R186 441,84. We are 8 
cents out. However, this is in fact not a mistake. Remember that when we worked out the monthly 
repayments, we rounded to the nearest cents and arrived at Rl 709,48. This was because one cannot 
make a payment for a fraction of a cent. Therefore, the rounding off error was carried through. That's 
why the two figures don't match exactly. In financial mathematics, this is largely unavoidable. 




4.5 Formula Sheet 




As an easy reference, here are the key formulae that we derived and used during this chapter. While 
memorising them is nice (there are not many), it is the application that is useful. Financial experts are 
not paid a salary in order to recite formulae, they are paid a salary to use the right methods to solve 
financial problems. 



Definitions 



EMCAN 



P Principal (the amount of money at the starting point of the calculation) 
i interest rate, normally the effective rate per annum 
n period for which the investment is made 



iT the interest rate paid T times per annum, i.e. iT ■■ 



Nominal Interest Rate 



r,:i 



4.5 



CHAPTER 4. FINANCE 



Equations 



EMCAO 



Present Value - simple 
Future Value - simple 
Solve for i 
Solve for n 



P(l + i.n) 



Tip 



Always keep the interest 
and the time period in 
the same units of time 
(e.g. both in years, or 
both in months etc.). 



Present Value - compound 
Future Value - compound 
Solve for i 
Solve for n 



P(l + *)" 



Chapter 4 


End of Chapter Exercises 



1. Thabo is about to invest his R8 500 bonus in a special banking product which will 
pay 1% per annum for 1 month, then 2% per annum for the next 2 months, then 3% 
per annum for the next 3 months, 4% per annum for the next 4 months, and 0% for 
the rest of the year. The bank is going to charge him R100 to set up the account. How 
much can he expect to get back at the end of the period? 

2. A special bank account pays simple interest of 8% per annum. Calculate the opening 
balance required to generate a closing balance of R5 000 after 2 years. 

3. A different bank account pays compound interest of 8% per annum. Calculate the 
opening balance required to generate a closing balance of R5 000 after 2 years. 

4. Which of the two answers above is lower, and why? 

5. 7 months after an initial deposit, the value of a bank account which pays compound 
interest of 7,5% per annum is R3 650,81. What was the value of the initial deposit? 

6. Thabani and Lungelo are both using UKZN Bank for their saving. Suppose Lungelo 
makes a deposit of X today at interest rate of i for six years. Thabani makes a deposit 
of 3X at an interest rate of 0,05%. Thabani made his deposit 3 years after Lungelo 
made his first deposit. If after 6 years, their investments are equal, calculate the value 
of i and find X. If the sum of their investment is R20 000, use the value of X to find 
out how much Thabani earned in 6 years. 

7. Sipho invests R500 at an interest rate of log(l,12) for 5 years. Themba, Sipho's sister 
invested R200 at interest rate i for 1 years on the same date that her brother made his 
first deposit. If after 5 years, Themba's accumulation equals Sipho's, find the interest 
rate i and find out whether Themba will be able to buy her favourite cell phone after 
10 years which costs R2 000. 

8. Calculate the real cost of a loan of R10 000 for 5 years at 5% capitalised monthly. 
Repeat this for the case where it is capitalised half yearly i.e. Every 6 months. 

9. Determine how long, in years, it will take for the value of a motor vehicle to decrease 
to 25% of its original value if the rate of depreciation, based on the reducing-balance 
method, is 21% per annum. 



51 



CHAPTER 4. FINANCE 4.5 



A" 1 ) More practice (Vj video solutions (?) or help at www.everythingmaths.co.z 



(1.)01e5 (2.)01e6 (3.) 01e7 (4.) 01e8 (5.) 01e9 (6.) 01ea 
(7.)01eb (8.)01ec (9.) 01 ed 



Factorising Cubic 
Polynomials 





5. 1 Introduction 




In Grades 10 and 1 1, you learnt how to solve different types of equations. Most of the solutions, relied 
on being able to factorise some expression and the factorisation of quadratics was studied in detail. 
This chapter focuses on the factorisation of cubic polynomials, that is expressions with the highest 
power equal to 3. 

See introductory video: VMgmo at www.everythingmaths.co.za 




5.2 The Factor Theorem 




The factor theorem describes the relationship between the root of a polynomial and a factor of the 
polynomial. 



DEFINITION: Factor Theorem 










For any polynomial, f(x), for al 
of f(x). Or, more concisely: 


values of a which satisfy f(a) = 





(x- 


- a) is a factor 




fix) = {x-a)q(x) 










where q(x) is a polynomial. 

In other words: If the remainder when dividing f(x) by (x — a) 

is a factor of f(x). 

So if /(-£) = 0, then (ax + b) is a factor of f{x). 


is 


zero 


then 


[x — a) 



Example 1: Factor Theorem 



56 



CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 5.2 

QUESTION 

Use the Factor Theorem to determine whether y — 1 is a factor of f(y) = 2y 4 + 3y 2 — by + 7. 

SOLUTION 



Step 1 : Determine how to approach the problem 

In order for y — 1 to be a factor, /(l) must be 0. 

Step 2 : Calculate /(l) 



f(y) = 2y i + 3y 2 -by + 7 

.-./(l) = 2(l) 4 + 3(l) 2 -5(l) + 7 

= 2+3-5+7 

= 7 

Step 3 : Conclusion 

Since /(l) ^ 0, y - 1 is not a factor of /(j/) = 2j/ 4 + 3j/ 2 - 5?/ + 7. 



Example 2: Factor Theorem 



QUESTION 



Using the Factor Theorem, verify that y + 4 is a factor of g(y) = by + 16y — \by + 8y + 16 



SOLUTION 



Step 1 : Determine how to approach the problem 

In order for y + 4 to be a factor, g(— 4) must be 0. 

Step 2 : Calculate /(l) 



g(y) = by" + 16j/ 3 - lby 2 + Sy + 16 

R.-.g(-A) = 5(-4) 4 + 16(-4) 3 -15(-4) 2 +8(-4) + 16 

= 5(256) + 16(-64) -15(16) + 8(-4) + 16 

= 1280 - 1024 - 240 -32+16 

= 



Step 3 : Conclusion 



5.3 



CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 



Since ff(-4) = 0, y + 4 is a factor of g(y) = 5?/ 4 + 16y 3 - 15y 2 + 8j/ + 16. 




5.3 Factorisation of Cubic 
Polynomials 




A cubic polynomial is a polynomial of the form 

ax + fox + ex + d 

where a is nonzero. We have seen in Grade 10 that the sum and difference of cubes is factorised as 
follows: 



(x + y)(x 2 -xy + y 2 ) = x 3 + y 3 



and 



(x - y)(x 2 + xy + y 2 ) = x 3 - y 3 
We also saw that the quadratic factor does not have real roots. 

There are many methods of factorising a cubic polynomial. The general method is similar to that used 
to factorise quadratic equations. If you have a cubic polynomial of the form: 

/(x) = ax + fox + ex + d 

then in an ideal world you would get factors of the form: 

(Ax + B)(Cx + D)(Ex + F). (5.1) 

But sometimes you will get factors of the form: 

{Ax + B)(Cx 2 +Ex + D) 

We will deal with simplest case first. When a = 1, then A = C = E = 1, and you only have to 
determine B, D and F. For example, find the factors of: 



In this case we have 

a = 1 

fo = -2 

c = -5 

d = 6 

The factors will have the general form shown in (5.1), with A = C = E = 1. We can then use values 
for a, fo, c and d to determine values for B, D and F. We can re-write (5.1) with A = C = E = 1 as: 



If we multiply this out we get: 

{x + B)(x + D)(x + F) 

We can therefore write: 



(x + B){x + D){x + F). 



(x + B)(x 2 + Dx + Fx + DF) 



x 3 + Dx 2 + Fx 2 + Bx 2 + DFx + BDx + BFx + BDF 
x 3 + (D + F + B)x 2 + {DF + BD + BF)x + BDF 



fo = 


-2 = D + F + B 


c = 


-5 = DF + BD + BF 


d = 


6 = BDF. 



(5.2) 
(5.3) 
(5.4) 



CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 5.3 



This is a set of three equations in three unknowns. However, we know that B, D and F are factors of 
6 because BDF = 6. Therefore we can use a trial and error method to find B, D and F. 
This can become a very tedious method, therefore the Factor Theorem can be used to find the factors 
of cubic polynomials. 



Example 3: Factorisation of Cubic Polynomials 



QUESTION 



Factorise f(x) = x 3 + x 2 — 9x — 9 into three linear factors. 



SOLUTION 



Step I : By trial and error using the factor theorem to find a factor 

Try 

/(l) = (l) 3 + (l) 2 - 9(1) -9 = 1 + 1-9-9 = -16 

Therefore (x — 1) is not a factor 
Try 

/(-l) = (-1) 3 + (-1) 2 9(-1)9 = 1 + 1 + 99 = 

Thus (x + 1) is a factor, because /(— 1) = 0. 

Now divide f(x) by (x + 1) using division by inspection: 
Write x 3 + x 2 -9x-9 = (x + 1)( ) 

The first term in the second bracket must be a; 2 to give x 3 if one works backwards. 
The last term in the second bracket must be —9 because +1 x — 9 = —9. 
So we have x 3 + x 2 - 9x - 9 = (x + l)(x 2 +?x - 9). 
Now, we must find the coefficient of the middle term (x). 

(+l)(x 2 ) gives the x 2 in the original polynomial. So, the coefficient of the x-term 
must be 0. 
So/(x) = (x + l)(x 2 -9). 



Step 2 : Factorise fully 

x 2 — 9 can be further factorised to (x — 3)(x + 3), 
and we are now left with /(x) = (x + l)(x — 3)(x + 3) 



In general, to factorise a cubic polynomial, you find one factor by trial and error. Use the factor theorem 
to confirm that the guess is a root. Then divide the cubic polynomial by the factor to obtain a quadratic. 
Once you have the quadratic, you can apply the standard methods to factorise the quadratic. 

For example the factors of x 3 — 2x 2 — 5x + 6 can be found as follows: There are three factors which 
we can write as 

(x — a)(x — b)(x — c). 



5.3 CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 



Example 4: Factorisation of Cubic Polynomials 



QUESTION 



Use the Factor Theorem to factorise 



SOLUTION 



Step 1 : Find one factor using the Factor Theorem 

Try 

/(l) = (l) 3 - 2(1) 2 - 5(1) + 6 = 1-2-5 + 6 = 
Therefore (x — 1) is a factor. 

Step 2 : Division by inspection 

x 3 -2x 2 -5x + 6 = (x- 1)( ) 

The first term in the second bracket must be x 2 to give x 3 if one works backwards. 
The last term in the second bracket must be —6 because — 1 x — 6 = +6. 
So we have x 3 - 2x 2 - 5x + 6 = (x - l)(x 2 +?x - 6). 
Now, we must find the coefficient of the middle term (x). 
(— l)(x 2 ) gives —x 2 . So, the coefficient of the x-term must be —1. 
So/(x) = (x- \){x 2 -x-6). 

Step 3 : Factorise fully 

x 2 — x — 6 can be further factorised to (x — 3)(x + 2), 
and we are now left with x 3 — 2x 2 — bx + 6 = (x — l)(x — 3)(x + 2) 



Exercise 5-1 



1 . Find the remainder when 4r 3 — Ax 2 + x — 5 is divided by (x + 1). 

2. Use the factor theorem to factorise x 3 — 3x 2 + 4 completely. 

3. f(x) = 2x 3 + x 2 -5x + 2 

(a) Find /(l). 

(b) Factorise f(x) completely 

4. Use the Factor Theorem to determine all the factors of the following expression: 

x" + x — YJx + 15 

5. Complete: If f(x) is a polynomial and p is a number such that f(p) = 0, then (x — p) is.. 



60 



CHAPTER 5. FACTOR1SING CUBIC POLYNOMIALS 



5.4 



More practice (►) video solutions (?) or help at www.everythingmaths.co.: 



(1.)01eh (2.)01ei (3.) Olej (4.) 01ek (5.) Olem 




5.4 Solving Cubic Equations 



Once you know how to factorise cubic polynomials, it is also easy to solve cubic equations of the kind 

ax + bx" + ex + d = 



Example 5: Solution of Cubic Equations 



QUESTION 


Solve 


6x 3 - 5x 2 - 17x + 6 = 0. 






SOLUTION 






Step 1 


: Find one factor using the Factor Theorem 

Try 








/(l) =6(1) 3 -5(1) 2 -17(1) + 6 = 6- 5- 17 + 6 = - 


-10 






Therefore (x — 1) is NOT a factor. 
Try 








/(2) = 6(2) 3 - 5(2) 2 - 17(2) + 6 = 48 - 20 - 34 + 6 = 


= 






Therefore (x — 2) IS a factor. 






Step 2 


: Division by inspection 

6x 3 - 5x 2 - 17x + 6 = (x - 2){ ) 
The first term in the second bracket must be 6a; 2 to give 6x 
wards. 

The last term in the second bracket must be —3 because —2 
So we have 6x 3 - 5x 2 - Ylx + 6 = (x - 2){6x 2 +?x - 3). 
Now, we must find the coefficient of the middle term (x). 
(— 2)(6x 2 ) gives — 12x 2 . So, the coefficient of the z-term mi 
So, 6a; 3 - 5x 2 - 17s + 6 = (x - 2)(6ir 2 + 7x - 3). 


3 if one 

x -3 = 

st be 7. 


works back- 
+6. 



(il 



5.4 CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 



Step 3 : Factorise fully 

6x 2 + 7x - 3 can be further factorised to (2x + 3)(3aj - 1), 
and we are now left with 6x 3 - 5x 2 - 17cc + 6 = (a; - 2)(2s + 3)(3x - 1) 



Step 4 : Solve the equation 



5x 2 - YJx + 6 = 



(»-2)(2w + 3)(3x- 1) 



2;i;- 3 
3 2 



Sometimes it is not possible to factorise the trinomial ("second bracket"). This is when the quadratic 
formula 



-b ± sJW- - iac 



2a 
can be used to solve the cubic equation fully. 
For example: 



Example 6: Solution of Cubic Equations 



QUESTION 



Solve for x: x 3 - 2x 2 - Gx + 4 = 0. 



SOLUTION 



Step 1 : Find one factor using the Factor Theorem 

Try 





/(1) = 


(l) 3 - 


-2(1) 2 


-6(1) 


Therefore (x 


- 1) is 


NOT a 


factor. 


Try 











/(2) = (2) 3 - 2(2) 2 - 6(2) + 4 = 8-8- 12 + 4 = -8 
Therefore (x — 2) is NOT a factor. 

/(-2) = (-2) 3 - 2(-2) 2 - 6(-2) + 4 = -8 -8 + 12 + 4 = 
Therefore (x + 2) IS a factor. 

Step 2 : Division by inspection 

x 3 -2x 2 -6x + 4 = (x + 2)( ) 

The first term in the second bracket must be x 2 to give x 3 . 
The last term in the second bracket must be 2 because 2x2 = +4. 
So we have x 3 - 2x 2 - 6x + 4 = (x + 2)(x 2 +?x + 2). 
Now, we must find the coefficient of the middle term (x). 
(2)(x 2 ) gives 2x 2 . So, the coefficient of the x-term must be —4. (2x 2 — Ax 2 



62 



CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 5.4 



-2x 2 ) 

So x 3 - 2x 2 - 6x + A = (x + 2)(x 2 - Ax + 2). 

x 2 — Ax + 2 cannot be factorised any further and we are now left with 

(x + 2)(x 2 -Ax + 2) = 



Step 3 : Solve the equation 



{x + 2)(x 2 - Ax + 2) = 

(x + 2) = or (x 2 -4x + 2) = 

Step 4 : Apply the quadratic formula for the second bracket 

Always write down the formula first and then substitute the values of a, b and c. 



X 


-b ± Vb 2 - Aac 
2a 






-(-4)±V(-4) 2 - 


-4(1)(2) 




2(1) 






A±V8 






2 






= 2±^2 




Step 5 : Final solutions 






x = — 2 or x = 


2±V2 





Chapter 5 



End of Chapter Exercises 



1 . Solve for x: x 3 + x 2 — 5x + 3 = 

2. Solve for y: y 3 - 3y 2 - l&y - 12 = 

3. Solve for m: m 3 — m 2 — Am — 4 = 

4. Solve for x: x 3 — x 2 = 3(3x + 2) Tip: Remove brackets and write as an equation 
equal to zero. 

5. Solve for x if 2x 3 - 3x 2 - 8x = 3 

6. Solve for x: 16(x + 1) = x 2 (x + 1) 

7. (a) Show that x - 2 is a factor of 3x 3 - llx 2 + 12x - 4 
(b) Hence, by factorising completely, solve the equation 

3x 3 - llx 2 + 12x - 4 = 

8. 2x 3 - x 2 - 2x + 2 = Q(x).(2x- 1) + i? for all values of x. What is the value of ffl. 

9. (a) Use the factor theorem to solve the following equation for m: 

8m 3 + 7m 2 - 17m + 2 = 
(b) Hence, or otherwise, solve for x: 



(53 



5.4 CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 

10. A challenge: 

Determine the values of p for which the function 

fix) = 3p — (3p — 7)x + 5x — 3 
leaves a remainder of 9 when it is divided by (x — p). 

UP) More practice CrJ video solutions CfJ or ne 'P at www.everythingmaths.co.za 

(1.)01en (2.)01ep (3.) 01eq (4.) 01er (5.)01es (6.) 01et 
(7.)01eu (8.) 01ev (9.) 01ew (10.) 01ex 



()i 



Functions and Graphs 





6. 1 Introduction 




In Grades 10 and 1 1 you learned about linear functions and quadratic functions, as well as the hyper- 
bolic functions and exponential functions and many more. In Grade 12 you are expected to demon- 
strate the ability to work with various types of functions and relations including inverses. In particular, 
we will look at the graphs of the inverses of: 

y = ax + q 

2 

y = ax 
y = ax; a > 

© See introductory video: VMgsz at www.everythingmaths.co.za 




6.2 Definition of a Function 



A function is a relation for which there is only one value of y corresponding to any value of x. We 
sometimes write y = fix), which is notation meaning 'y is a function of x'. This definition makes 
complete sense when compared to our real world examples — each person has only one height, so 
height is a function of people; on each day, in a specific town, there is only one average temperature. 

However, some very common mathematical constructions are not functions. For example, consider the 
relation x 2 + y 2 = 4. This relation describes a circle of radius 2 centred at the origin, as in Figure 6.1 . 



If we let x = 0, we see that y 2 = 4 and thus either y 
which are possible for the same x value, the relation x 2 



2 or y = —2. Since there are two y values 
- y 2 = 4 is not the graph a function. 



There is a simple test to check if a relation is a function, by looking at its graph. This test is called the 
vertical line test. If it is possible to draw any vertical line (a line of constant x) which crosses the graph 
of the relation more than once, then the relation is not a function. If more than one intersection point 
exists, then the intersections correspond to multiple values of y lor a single value of x. 

We can see this with our previous example of the circle by looking at its graph again in Figure 6.1. 

We see that we can draw a vertical line, for example the dotted line in the drawing, which cuts the 
circle more than once. Therefore this is not a function. 



Exercise 6-7 



1 . State whether each of the following equations are functions or not: 

(a) x + y = 4 

(b) y = f 

(c) y = T 



65 



6.3 



CHAPTER 6. FUNCTIONS AND GRAPHS 




Figure 6.1: Graph of x 2 



(d) 



If 



2. The table gives the average per capita income, d, in a region of the country as a function of u, 
the percentage unemployed. Write down the equation to show that income is a function of the 
percent unemployed. 



u 


1 


2 


3 


4 


d 


22500 


22000 


21500 


21000 



t/Vj More practice CrJ video solutions ( 9 J or help at www.everythingmaths.co.za 



(1.)0ley (2.)01ez 




6.3 Notation Used for Functions 




In Grade 1 you were introduced to the notation used to name a function. In a function y = f(x), y is 
called the dependent variable, because the value of y depends on what you choose as x. We say x is 
the independent variable, since we can choose x to be any number. Similarly if g(t) = 2t + 1, then t 
is the independent variable and g is the function name. If f(x) = 3x — 5 and you are ask to determine 
/(3), then you have to work out the value for f(x) when x = 3. For example, 



/(3) 



3x — 

3(3) 
4 



(i(i 




CHAPTER 6. FUNCTIONS AND GRAPHS 6.4 



6.4 Graphs of Inverse Functions 



In earlier grades, you studied various types of functions and understood the effect of various parameters 
in the general equation. In this section, we will consider inverse functions. 

An inverse function is a function which does the reverse of a given function. More formally, if / is a 
function with domain X, then / _1 is its inverse function if and only if for every ielwe have: 

r 1 u(x)) = x (6.D 

A simple way to think about this is that a function, say y = f(x), gives you a y-value if you substitute an 
i-value into f(x). The inverse function tells you tells you which x-value was used to get a particular 
j/-value when you substitute the y-value into f~ 1 (y). There are some things which can complicate 
this - for example, for the sin function there are many x-values that give you a peak as the function 
oscillates. This means that the inverse of the sin function would be tricky to define because if you 
substitute the peak j/-value into it you won't know which of the i-values was used to get the peak. 



y = f(x) we have a function 
g/i = f{xi) we substitute a specific x-value into the function to get a specific j/-value 

consider the inverse function 
x = f' 1 {y) 
x = f~ (y) substituting the specific y-value into the inverse should return the specific rc-value 

= F\yi) 

= X! 

This works both ways, if we don't have any complications like in the case of the sin function, so we 
can write: 

r 1 (f(x)) = f(r 1 (x)) = x (6.2) 

(x - 2) 
For example, if the function x — > 3x + 2 is given, then its inverse function is x — > - — - — -. This is 

usually written as: 

/ : x^3x + 2 (6.3) 

r 1 ■ »-^ (6.4) 

The superscript " — 1" is not an exponent. 

If a function / has an inverse then / is said to be invertible. 

If / is a real-valued function, then for / to have a valid inverse, it must pass the horizontal line test, 
that is a horizontal line y = k placed anywhere on the graph of / must pass through / exactly once 
for all real k. 

It is possible to work around this condition, by defining a multi-valued function as an inverse. 

If one represents the function / graphically in a zy-coordinate system, the inverse function of the 
equation of a straight line, f 1 , is the reflection of the graph of / across the line y = x. 

Algebraically, one computes the inverse function of / by solving the equation 

y = f(x) 

for x, and then exchanging y and x to get 

y = /^M 

See video: VMguj at www.everythingmaths.co.za 



6.4 



CHAPTER 6. FUNCTIONS AND GRAPHS 



Inverse Function ofy = ax + q 



EMCAX 



The inverse function of y = ax + q is determined by solving for x as: 



(6.5) 
(6.6) 

(6.7) 
(6.8) 



Therefore the inverse of y ■■ 

The inverse function of a straight line is also a straight line, except for the case where the straight line 
is a perfectly horizontal line, in which case the inverse is undefined. 

For example, the straight line equation given by y = 2x — 3 has as inverse the function, y — |x + §. 
The graphs of these functions are shown in Figure 6.2. It can be seen that the two graphs are reflections 
of each other across the line y = x. 



y 


- 


ax + q 


ax 


= 


y-q 


X 


= 


y-q 

a 




= 


i q 
-v-- 

a a 


ax + q\iy = -X 


_ £ 

a 





r\x) 




Figure 6.2: The graphs of the function f(x) = 2x — 3 and its inverse / i (x) = \x+ ^. The line y ■■ 
is shown as a dashed line. 



Domain and Range 

We have seen that the domain of a function of the form y = ax + q is {x : x e M} and the range is 
{y : y 6 M}. Since the inverse function of a straight line is also a straight line, the inverse function will 
have the same domain and range as the original function. 



Intercepts 

The general form of the inverse function of the form y = ax + q is y = -x — -. 



(i.s 



CHAPTER 6. FUNCTIONS AND GRAPHS 6.4 



By setting i = Owe have that the ^-intercept is y int = — J. Similarly, by setting y = we have that 
the x-intercept is Xmt = 1- 

It is interesting to note that if f(x) = ax + q, then f 1 {x) = -x — 2 and the j/-intercept of f(x) is the 
z-intercept of / _1 (x) and the x-intercept of f(x) is the j/-intercept of f~ L (x). 



Exercise 6-2 



1 . Given f(x) = 2x - 3, find f^{x) 

2. Consider the function f(x) = 3x — 7. 

(a) Is the relation a function? 

(b) If it is a function, identify the domain and range. 

3. Sketch the graph of the function f(x) = 3x — 1 and its inverse on the same set of axes. 

4. The inverse of a function is f^ 1 (x) = 2x — 4, what is the function /(a;)? 

(f^j More practice Crj video solutions Cf) or help at www.everythingmaths.co.za 
(1.)01fO (2.) 01 f1 (3.)01f2 (4.) 01 f3 

Inverse Function ofy = ax 2 wemcay 

The inverse relation, possibly a function, of y = ax 2 is determined by solving for x as: 



y = ax (6.9) 

x 2 ¥■ (6.10) 

±J- (6.11) 



(i'l 



6.4 



CHAPTER 6. FUNCTIONS AND GRAPHS 



f(x)=x 2 




r\*) = sfc 



r i (x) = -^ 



Figure 6.3: The function f(x) = x 2 and its inverse / L (x) = ±^/x. The line y = x is shown as a 
clashed line. 



We see that the inverse relation of y 



is not a function because it fails the vertical line test. If 



we draw a vertical line through the graph of f~ 1 (x) = ±^/x, the line intersects the graph more than 
once. There has to be a restriction on the domain of a parabola for the inverse to also be a function. 
Consider the function f(x) = —x 2 + 9. The inverse of / can be found by writing f(y) = x. Then 



x = —y + 9 
y = 9 — x 
y = ±\/9-x 



If we restrict the domain of f(x) to be x > 0, then \/9 — x is a function. If the restriction on the 
domain of / is x < then —\/9 — x would be a function, inverse to /. 

© See video: VMgvs at www.everythingmaths.co.za 



Exercise 6-3 



1 . The graph of / Ms shown. Find the equation of /, given that the graph of / is a parabola. (Do 
not simplify your answer) 




CHAPTER 6. FUNCTIONS AND GRAPHS 6.4 

2. f(x) = 2x 2 . 

(a) Draw the graph of / and state its domain and range. 

(b) Find / _1 and, if it exists, state the domain and range. 

(c) What must the domain of / be, so that / _1 is a function ? 



3. Sketch the graph of x = — -y/10 — y 2 . Label a point on the graph other than the intercepts with 
the axes. 



4. (a) Sketch the graph of y = x 2 labelling a point other than the origin on your graph. 

(b) Find the equation of the inverse of the above graph in the form y = . . . 

(c) Now sketch the graph of y = </x. 

(d) The tangent to the graph of y = sfx at the point A(9; 3) intersects the z-axis at B. Find the 
equation of this tangent and hence or otherwise prove that the y-axis bisects the straight 
line AB. 

5. Given: g(x) = — 1 + </x, find the inverse of g(x) in the form g^ 1 {x) = . . . 



UP) More practice CrJ video solutions Cf) or help at www.everythingmaths.co.za 



(1.)01f4 (2.) 01f5 (3.)01f6 (4.) 01f7 (5.) 021f 



Inverse Function ofy = a x wemcaz 



The inverse function of y = a x is determined by solving for x as follows: 



y = a (6.12) 

log(j/) = log(a*) (6.13) 

= xlog(a) (6.14) 

log(y) 



log(a) 
The inverse of y = 1CT is x = log(y) Therefore, if f(x) = lO 1 ', then /(x) _1 = log(x). 



(6.15) 



71 



6.4 



CHAPTER 6. FUNCTIONS AND GRAPHS 



fix) = 10* 




r i (x) = io g (x) 



Figure 6.4: The function f(x) = 1(F and its inverse / 1 (x) = log(x). The line y = x is shown as a 
clashed line. 



The exponential function and the logarithmic function are inverses of each other; the graph of the one 
is the graph of the other, reflected in the line y = x. The domain of the function is equal to the range 
of the inverse. The range of the function is equal to the domain of the inverse. 



Exercise 6-4 



1 . Given that f(x) = (\) x , sketch the graphs of / and / x on the same system of axes indicating 
a point on each graph (other than the intercepts) and showing clearly which is / and which is 

r 1 . 

2. Given that /(x) = 4 _x , 

(a) Sketch the graphs of /and f~ l on the same system of axes indicating a point on each graph 
(other than the intercepts) and showing clearly which is / and which is / _1 . 

(b) Write / _1 in the form y = . . . 

3. Given g(x) = — 1 + \[x, find the inverse of g(x) in the form g^ 1 (x) = . . . 

4. (a) Sketch the graph of y = x 2 , labelling a point other than the origin on your graph. 

(b) Find the equation of the inverse of the above graph in the form y = . . . 

(c) Now, sketch y = *Jx. 

(d) The tangent to the graph of y = *Jx at the point A(9; 3) intersects the x-axis at B. Find the 
equation of this tangent, and hence, or otherwise, prove that the j/-axis bisects the straight 
line AB. 



72 



CHAPTER 6. FUNCTIONS AND GRAPHS 



6.4 



A" 1 ) More practice (►) video solutions (fj or help at www.everythingmaths.co.: 



(1.)01f8 (2.) 01 f9 (3.) 01fa (4.) 021 § 



Chapter 6 



End of Chapter Exercises 



1 . Sketch the graph of x = — -^/10 — y 2 . Is this graph a function? Verify your answer. 

2- /(*) = -^=, 

x — 5 

(a) determine the j/-intercept of f(x) 

(b) determine x if f(x) = — 1. 

3. Below, you are given three graphs and five equations. 

Graph 1 Graph 2 Graph 3 



I MP x 





(a) j/ = log 3 x 

(b) 3/ = - log 3 x 

(c) 2/ = log 3 (-x) 

(d) y = 3-" 

(e) y = T 

Write the equation that best matches each graph. 

4. Given g(x) = — 1 + y^< find the inverse of g(x) in the form g^ 1 (x) = . . . 

5. Consider the equation h(x) = 3 X 

(a) Write down the inverse in the form /i _1 (x) = . . . 

(b) Sketch the graphs of h(x) and h~ 1 (x) on the same set of axes, labelling the 
intercepts with the axes. 

(c) For which values of x is h^ 1 (x) undefined ? 

6. (a) Sketch the graph of y = 2x 2 + 1, labelling a point other than the origin on your 

graph. 



6.4 CHAPTER 6. FUNCTIONS AND GRAPHS 



(b) Find the equation of the inverse of the above graph in the form y = . . . 

(c) Now, sketch y = s/x. 

(d) The tangent to the graph of y = ^/x at the point A(9; 3) intersects the x-axis at 
B. Find the equation of this tangent, and hence, or otherwise, prove that the 
y-axis bisects the straight line AB. 



More practice (►) video solutions (0 J or help at www.everythingmaths.co.za 
(1.)01fb (2.)01fc (3.) Olfd (4.)01fe (5.) 01ff (6.) Olfg 



7i 



Differential Calculus 





7.1 Introduction 




Calculus is one of the central branches of mathematics and was developed from algebra and geometry. 
Calculus is built on the concept of limits, which will be discussed in this chapter. Calculus consists 
of two complementary ideas: differential calculus and integral calculus. We will only be dealing with 
differential calculus in this text and will explore how it can be used to address mathematical problems 
like optimisation and finding rates of change. 

© See introductory video: VMgxl at www.everythingmaths.co.za 




7.2 Limits 




The Tale of Achilles and the Tortoise 



EMCBC 



One of Zeno's paradoxes can be summarised as: 



Achilles and a tortoise agree to a race, but the tortoise is unhappy because Achilles is very 
fast. So, the tortoise asks Achilles for a head-start. Achilles agrees to give the tortoise a 
1000m head start. Does Achilles overtake the tortoise? 



We know how to solve this problem. We start by writing: 

xa = 11 At (7.1) 

x t = 1000m + v t t (7.2) 



FACT 



Zeno (circa 490 BC - 
circa 430 BC) was a pre- 
Socratic Greek philoso- 
pher of southern Italy 
who is famous for his 
paradoxes. 



where 



xa distance covered by Achilles 

va Achilles' speed 

t time taken by Achilles to overtake tortoise 

x t distance covered by the tortoise 

Vt the tortoise's speed 



If we assume that Achilles runs at 2m ■ s _1 and the tortoise runs at 0,25 1 
overtake the tortoise when both of them have covered the same distance, 
overtakes the tortoise at a time calculated as: 



is 1 then Achilles will 
This means that Achilles 



75 



7.2 CHAPTER 7. DIFFERENTIAL CALCULUS 



xa = xt (7.3) 

VAt = 1000 + v t t (7.4) 

(2m-s _1 )i = 1000m + (0,25 m-s _1 )t (7.5) 

(2m s _1 -0,25m -s _1 )i = 1000m (7.6) 

t = ™^ (7.7, 
l4m- s- 1 



1000 m 
(4) (1000) 



(7.8) 



(7.9) 



= <%>. (7,0, 

= 571^s (7.11) 

However, Zeno (the Greek philosopher who thought up this problem) looked at it as follows: Achilles 

takes 

1000 m 

t = r = 500 s 

2m ■ s -1 

to travel the 1 000 m head start that the tortoise had. However, in these 500 s, the tortoise has travelled 

a further 

x = (500s)(0,25ms _1 ) = 125 m. 

Achilles then takes another 

125 m 

t = ~ r = 62 ' 5s 

2 m -s- 1 

to travel the 125 m. In these 62,5 s, the tortoise travels a further 

x = (62,5s)(0,25m-s _1 ) = 15,625 m. 
Zeno saw that Achilles would always get closer but wouldn't actually overtake the tortoise. 



Sequences, Series and Functions m emcbd 



So what does Zeno, Achilles and the tortoise have to do with calculus? 
Well, in Grades 10 and 1 1 you studied sequences. For the sequence 

o-i- 2 - 3 - 4 - 

'2'3'4'5' 

which is defined by the expression 

On = 1 

n 
the terms get closer to 1 as n gets larger. Similarly, for the sequence 

' 2'3'4' 5' 

which is defined by the expression 

1 

On = — 

11 

the terms get closer to as n gets larger. We have also seen that the infinite geometric series can have 
a finite total. The infinite geometric series is 



Soo = > Oi.r 1 = for — 1 < r < 1 



l: 1 



76 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.2 



where a\ is the first term of the series and r is the common ratio. 

We see that there are some functions where the value of the function gets close to or approaches a 
certain value. 

Similarly, for the function: 

x 2 + 4x - 12 



x + 6 



The numerator of the function can be factorised as: 



Qr + 6)(x-2) 
x + 6 



Then we can cancel the x + 6 from numerator and denominator and we are left with: 



However, we are only able to cancel the x + 6 term if x =£ —6. If x = —6, then the denominator 
becomes and the function is not defined. This means that the domain of the function does not 
include x = —6. But we can examine what happens to the values for y as x gets closer to —6. These 
values are listed in Table 7.1 which shows that as x gets closer to —6, y gets close to 8. 



Table 7.1 : Values for the function 



(x + 6)(x-2) 
x + 6 



as x gets close to -6. 



X 


_ (x + 6)(x-2) 

y x+6 


-9 


-li 


-8 


-10 


-7 


-9 


-6.5 


-8.5 


-6.4 


-8.4 


-6.3 


-8.3 


-6.2 


-8.2 


-6.1 


-8.1 


-6.09 


-8.09 


-6.08 


-8.08 


-6.01 


-8.01 


-5.9 


-7.9 


-5.8 


-7.8 


-5.7 


-7.7 


-5.6 


-7.6 


-5.5 


-7.5 


-5 


-7 


-4 


-6 


-3 


-5 



The graph of this function is shown in Figure 7.1. The graph is a straight line with slope 1 and intercept 
—2, but with a hole at x = —6. 



7.2 



CHAPTER 7. DIFFERENTIAL CALCULUS 




Figure 7.1: Graph of y ■■ 



(ai + 6)(x-2) 



Limits 



EMCBE 



We can now introduce a new notation. For the function 



(x + 6)(x--2) 
x + 6 



we can write: 



fan (3: + 6)( *- 2) = -8. 



This is read: the limit of < x + 6 >(^ 2 > as x tenc j s to _6 ; s _g_ 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.2 



Activity: 



Limits 



If /(a 



1, determine: 



/(-o.i) 




/(-0.05) 




/(-0.04) 




/(-0.03) 




/(-0.02) 




/(-o.oi) 




/(O.OO) 




/(O.OI) 




/(0.02) 




/(0.03) 




/(o.oi) 




/(0.05) 




/(o.i) 





What do you notice about the value of f(x) as ie gets close to 0? 



Example 1: Limits Notation 



QUESTION 


Summarise the following 
the function 


situation 


by 


using limit notation: As x gets close to 1, the value of 

y = x + 2 


gets close to 3. 








SOLUTION 








This is written as: 






lim x + 2 = 3 

X->1 


in limit notation. 











We can also have the situation where a function has a different value depending on whether x ap- 
proaches from the left or the right. An example of this is shown in Figure 7.2. 



7.2 



CHAPTER 7. DIFFERENTIAL CALCULUS 



H h 



H 1 1 fill- 




7 x 



Figure 7.2: Graph of y = -. 



As x — y from the left, y = - approaches — oo. As x — > from the right, y = - approaches +oo. This 
is written in limits notation as: 



for x approaching from the left and 



lim — = — oo 

x^O- X 



lim — = oo 
x-+o+ x 



for x approaching from the right. You can calculate the limit of many different functions using a set 
method. 

Method: 

Limits: If you are required to calculate a limit like lim x ^„ then: 

1 . Simplify the expression completely. 

2. If it is possible, cancel all common terms. 

3. Let a; approach a. 



Example 2: Limits 



QUESTION 



Determine 



lim 10 



80 



CHAPTER 7. DIFFERENTIAL CALCULUS 7.2 



SOLUTION 



Step 1 : Simplify the expression 

There is nothing to simplify. 

Step 2 : Cancel all common terms 

There are no terms to cancel. 

Step 3 : Let x — > 1 and write final answer 



lim 10 = 10 

x-tl 



Example 3: Limits 



QUESTION 



Determine 



lim c 

x^2 



SOLUTION 



Step 1 : Simplify the expression 

There is nothing to simplify. 

Step 2 : Cancel all common terms 

There are no terms to cancel. 

Step 3 : Let x — > 2 and write final answer 



lim x = 2 

x^2 



81 



7.2 CHAPTER 7. DIFFERENTIAL CALCULUS 



Example 4: Limits 



QUESTION 


Determine 


lim - 


2 - 100 






x-10 




SOLUTIO\ 








Step 1 : 


Simplify the expression 

The numerator can be factorised 








x 2 - 100 (x + 10)(x- 


10) 




x-10 


x-10 




Step 2 : 


Cancel all common terms 








(x — 10) can be cancelled from 


the numerator and denominator. 




(x + 10)(a;- 
x- 10 


10 ». I + io 


Step 3 : 


Let x — > 10 and write final answer 






x 2 - 
lim 

£-►10 x - 


H?-» 







Average Gradient and Gradient at a Point W emcbf 



In Grade 10 you learnt about average gradients on a curve. The average gradient between any two 
points on a curve is given by the gradient of the straight line that passes through both points. In Grade 
1 1 you were introduced to the idea of a gradient at a single point on a curve. We saw that this was 
the gradient of the tangent to the curve at the given point, but we did not learn how to determine the 
gradient of the tangent. 

Now let us consider the problem of trying to find the gradient of a tangent t to a curve with equation 
y = /(x) at a given point P. 




82 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.2 



We know how to calculate the average gradient between two points on a curve, but we need two 
points. The problem now is that we only have one point, namely P. To get around the problem we 
first consider a secant to the curve that passes through point P and another point on the curve Q, 
where Q is an arbitrary distance (h) from P, as shown in the figure. We can now find the average 
gradient of the curve between points P and Q. 



f(a + h) 
f(a) 




secant 



If the x-coordinate of P is a + h, then the ^-coordinate is /(a + h). Similarly, if the z-coordinate of Q 
is a, then the ^-coordinate is f(a). If we choose a + h as x-i and a as xi, then: 

2A = /0) 
y-i = f(a + h). 
We can now calculate the average gradient as: 

2/2-2/1 f(a + h) - f(a) 



(a + h) — a 

f(a + h)-f(a) 

h 



(7.12) 

(7.13) 



Now imagine that Q moves along the curve toward P. The secant line approaches the tangent line as 
its limiting position. This means that the average gradient of the secant approaches the gradient of the 
tangent to the curve at P. In (??) we see that as point Q approaches point P, h gets closer to 0. When 
h = 0, points P and Q are equal. We can now use our knowledge of limits to write this as: 



gradient at P = lim 

h->0 



f(g + h)-f(h) 
h 



(7.14) 



and we say that the gradient at point P is the limit of the average gradient as Q approaches P along 
the curve. 

© See video: VMgye at www.everythingmaths.co.za 



Example 5: Limits 



QUESTION 



For the function f(x) = 2x 2 — 5x, determine the gradient of the tangent to the curve at the 



S:{ 



7.2 CHAPTER 7. DIFFERENTIAL CALCULUS 



point x = 2. 



SOLUTION 



Step 1 : Calculating the gradient at a point 

We know that the gradient at a point x is given by: 

lim f{x + k) - f{x) 
h-»0 h 

In our case x = 2. It is simpler to substitute x = 2 at the end of the calculation. 
Step 2 : Write f(x + h) and simplify 

f(x + h) = 2{x + h) 2 -5(x + h) 

= 2(x 2 + 2xh + h 2 ) - 5x - 5h 
= 2x 2 + 4xh + 2h 2 -5x-5h 

Step 3 : Calculate limit 

f(x + h)-f(x) 2x 2 + Axh + 2h 2 - 5x - 5h - (2x 2 - 5x) _ 

fc->o h h 

,. 2x 2 + 4z/i + 2/i 2 - 5k - 5/i - 2x 2 + 5x 

= lim 

d-»o h 

&xh + 2/i 2 - 5/i 



lim 
h-X) 

lim 



h-»o /i 

h(4x + 2h-5) 

lim Ax + 2h - 5 



= 4a; - 5 

Step 4 : Calculate gradient at x = 2 

4x - 5 = 4(2) -5 = 3 

Step 5 : Write the final answer 

The gradient of the tangent to the curve f(x) = 2x 2 — 5x at x = 2 is 3. This is 
also the gradient of the curve at x = 2. 



Example 6: Limits 



QUESTION 



For the function f(x) = 5x 2 - 4x + 1, determine the gradient of the tangent to curve at the 



<sl 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.2 



point x = a. 



SOLUTION 



Step 1 : Calculating the gradient at a point 

We know that the gradient at a point x is given by: 



lim 



f(x + h)-f(x) 



In our case x = a. It is simpler to substitute x = a at the end of the calculation. 
Step 2 : Write f(x + ft) and simplify 

f(x + h) = 5(x + hf - A(x + ft) + 1 

= 5(x 2 + 2xft + ft 2 ) -4x-4ft + l 
= 5x 2 + lOxh + 5ft 2 - 4x - 4ft + 1 



Step 3 : Calculate limit 

/(x + ft)-/(x) 



lim 
7i-t0 



ft 



5x 2 + lOxft + 5ft 2 - 4x - 4ft + 1 - (5x 2 - 4x + 1) 
ft 
5x 2 + lOxft + 5ft 2 - 4x - 4ft + 1 - 5x 2 + 4x - 1 



lim 
h -to 

lim 
ft -to 

lim 

ft-to ft 

lim lOx + 5ft - 4 
10x-4 



10xft + 5ft 2 -4ft 



ft(10x + 5ft-4) 



Step 4 : Calculate gradient at x = a 

10x - 4 = 10a - 4 

Step 5 : Write the final answer 

The gradient of the tangent to the curve /(x) = 5x 2 — Ax + 1 at x = 1 is 10a — 4. 



Exercise 7-1 



Determine the following 
1. 

2. 



lim 



x 2 -9 



a-t3 X + 3 



lim 



x + 3 



x-t3 X 2 + 3x 



85 



7.3 



CHAPTER 7. DIFFERENTIAL CALCULUS 



3. 



lim 



3z - 4x 



lim 



2 3-x 

x 2 -x- 12 



+ 4 x - 4 



lim 3z + — 

x^2 3x 



Aj More practice (►) video solutions Cf) or help at www.everythingmaths 



(l.)01fh (2.)01fi (3.) 01 fj (4.) 01 fk (5.) 01fm 




7.3 Differentiation from First 
Principles 




The tangent problem has given rise to the branch of calculus called differential calculus and the 
equation: 

Um f(x + h)- f(x) 
fc-MJ h 

defines the derivative of the function f(x). Using (7.15) to calculate the derivative is called finding 
the derivative from first principles. 



Tip 



Though we choose to 
use a fractional form of 
representation, jg is a 
limit and is not a frac- 
tion, i.e. §^ does not 

' dx 

mean dy -f- dx. =^ 

" dx 

means y differentiated 
with respect to x. Thus, 
^ means p differenti- 
ated with respect to x. 



ator", applied to some 
function of x. 



DEFINITION: Derivative 
The derivative of a function f(x) is written as f'(x) and is defined by: 

f{x) = lim lk+K >-/w 



(7.15) 



There are a few different notations used to refer to derivatives. If we use the traditional notation 
y = f{x) to indicate that the dependent variable is y and the independent variable is x, then some 
common alternative notations for the derivative are as follows: 



/'(*) 



.'/ 



dy _ df_ 
dx dx 



dx 



f{x) = Df(x) = D x f(x) 



The symbols D and 4- are called differential operators because they indicate the operation of differ- 
entiation, which is the process of calculating a derivative. It is very important that you learn to identify 
these different ways of denoting the derivative, and that you are consistent in your usage of them when 
answering questions. 

See video: VMgyp at www.everythingmaths.co.za 



Mi 



CHAPTER 7. DIFFERENTIAL CALCULUS 7.3 



Example 7: Derivatives - First Principles 



QUESTION 



Calculate the derivative of g(x) = x — 1 from first principles. 



SOLUTION 



Step I : Calculate the gradient at a point 

We know that the gradient at a point x is given by: 

// x ,- 9(x + h) -g(x) 
q (x) = urn — 7 ~^- 

Step 2 : Write g(x + ft) and simplify 

g(x + ft) = x + ft — 1 
Step 3 : Calculate limit 



1 1 x ,. g(x + h) - g(x) 

q (x) = lim — i -^- 



x + h - 


-1- 


(x-l) 


h-¥0 


ft 




x + h- 


- 1 - 


x + 1 


h-K3 


ft 




lim — 

h^O h 






lim 1 






1 







Step 4 : Wrife f/ie final answer 

The derivative g'(a;) of g(x) = x — 1 is 1. 



Exercise 7-2 



1 . Given g(x) = — x 2 

, , , . o(a: + ft) — o(a;) 

(a) determine ^ / — ! ^— L 

h 

(b) hence, determine 

Um g(x + h)-g(x) 

h-»0 ft 

(c) explain the meaning of your answer in (b). 

2. Find the derivative of f(x) = —2x 2 + 3x using first principles. 



87 



7.4 



CHAPTER 7. DIFFERENTIAL CALCULUS 



3. Determine the derivative of f(x) 



x-2 



using first principles. 



4. Determine /'(3) from first principles if f(x) = —5a; 2 . 

5. If h(x) = 4a: 2 — Ax, determine h'(x) using first principles. 



CjX*y More practice f ►) video solutions f? J or help at www.everythingmaths.co.za 



(1.)01fn (2.)01fp (3.) 01fq (4.) 01fr (5.) 01fs 




7.4 Rules of Differentiation 




Calculating the derivative of a function from first principles is very long, and it is easy to make mistakes. 
Fortunately, there are rules which make calculating the derivative simple. 



Activity: 



Rules of Differentiation 



From first principles, determine the derivatives of the following: 

1. f(x) = b 

2. f(x) = x 

3. f(x) = x 2 

4. f(x) = x 3 

5. f(x) = l/x 

You should have found the following: 



/(*) 


/'(*) 


b 





X 


1 


x 1 


2x 


x :i 


3x 2 


l/x = x~ L 


—x 



If we examine these results we see that there is a pattern, which can be summarised by: 



dx 



(X ) = nx 



(7.16) 



There are two other rules which make differentiation simpler. For any two functions f(x) and g(x): 

d 



dx 



lf(x)±g(x)} = f'(x)±g'(x) 



This means that we differentiate each term separately. 

The final rule applies to a function f(x) that is multiplied by a constant k. 

^ x [k.f(x)} = kf'(x) 
See video: VMhgb at www.everythingmaths.co.za 



(7.17) 



(7.18) 



,ss 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.4 



Example 8: Rules of Differentiation 



QUESTION 


Determine the derivative of x — 1 using the rules of diffei 


entiation. 


SOLUTION 




Step 1 


Identify the rules that will be needed 

We will apply two rules of differentiation: 

0* t n \ n — 1 
— — (x I = nx 
dx 

and 






i\m- a w = i\m- 


■5W*M 


Step 2 


Determine the derivative 

In our case f(x) = x and g(x) = 1. 

fix) = 1 

and 

g'(x) = 




Step 3 


Write the final answer 






The derivative of x — 1 is 1 which is the same result as was obtained earlier, from 




first principles. 








Given two functions, f(x) and g(x) and constant b, n and k, we know that: 



^6 = 



£(x")=nx-i 



dx V J I dx 



dx w + 9) dx + dx 



89 



7.5 



CHAPTER 7. DIFFERENTIAL CALCULUS 



Exercise 7-3 




1. Find/'(x)if/(x) = ^- 


5x + 6 
-2 


2. Find f'(y) if f(y) = rfj. 


3. Kndf'(z)\ff(z) = (z-l)(z + l). 


4. Determine f if y = ** + 2 ^ ~ 3 . 


5. Determine the derivative of y = ya? -\ 

3a; 



l\n More practice (►) video solutions (9j or help at www.everythingmaths 



(1.) 01ft (2.)01fu (3.)01fv (4.)01fw (5.)01fx 



7.5 Applying Differentiation to 
Draw Graphs 



EMCB] 



Thus far we have learnt about how to differentiate various functions, but I am sure that you are be- 
ginning to ask, What is the point of learning about derivatives? Well, we know one important fact 
about a derivative: it is a gradient. So, any problems involving the calculations of gradients or rates of 
change can use derivatives. One simple application is to draw graphs of functions by firstly determine 
the gradients of straight lines and secondly to determine the turning points of the graph. 



Finding Equations of Tangents to Curves 



EMCBK 



In Section 7.2.4 we saw that finding the gradient of a tangent to a curve is the same as finding the 
gradient (or slope) of the same curve at the point of the tangent. We also saw that the gradient of a 
function at a point is just its derivative. 

Since we have the gradient of the tangent and the point on the curve through which the tangent passes, 
we can find the equation of the tangent. 



Oil 



CHAPTER 7. DIFFERENTIAL CALCULUS 


7.5 










Example 9: Finding the Equation of a Tangent to a Curve 




QUESTION 





Find the equation of the tangent to the curve y = x 2 at the point (1; 1) and draw both 
functions. 



SOLUTION 



Step 1 : Determine what is required 

We are required to determine the equation of the tangent to the curve defined 
by y = x 2 at the point (1; 1). The tangent is a straight line and we can find the 
equation by using derivatives to find the gradient of the straight line. Then we 
will have the gradient and one point on the line, so we can find the equation 
using: 

y — j/i = m(x — xi) 

from Grade 1 1 Coordinate Geometry. 

Step 2 : Differentiate the function 

Using our rules of differentiation we get: 

y = 2x 

Step 3 : Find the gradient at the point (1; 1) 

In order to determine the gradient at the point (1: 1), we substitute the x-value 
into the equation for the derivative. So, y' at x = 1 is: 

m = 2(1) = 2 
Step 4 : Find the equation of the tangent 



l-Vi 


= 


m{x — Xi) 


0-1 


= 


(2)(x-l) 


y 


= 


2x - 2 + 1 


y 


= 


2x- 1 



Step 5 : Write the final answer 

The equation of the tangent to the curve defined by y = x 2 at the point (1; 1) is 
y = 2x - 1. 

Step 6 : Sketch both functions 



91 



7.5 



CHAPTER 7. DIFFERENTIAL CALCULUS 



-\ 1 1 h 

-4 -3 -2 -1 



y = x 




y = 2x - 1 



Curve Sketching 



EMCBL 



Tip 


If X 


= a is 


a turning 


point of f(x), 


then: 




/'(a) = 


= 


This 


means 


that the 


derivative is 


at a 


turnir 


g point. 





Differentiation can be used to sketch the graphs of functions, by helping determine the turning points. 
We know that if a graph is increasing on an interval and reaches a turning point, then the graph will 
start decreasing after the turning point. The turning point is also known as a stationary point because 
the gradient at a turning point is 0. We can then use this information to calculate turning points, by 
calculating the points at which the derivative of a function is 0. 

Take the graph of y = x 2 as an example. We know that the graph of this function has a turning point 
at (0,0), but we can use the derivative of the function: 

y = 2x 

and set it equal to to find the z-value for which the graph has a turning point. 

2x = 

x = 

We then substitute this into the equation of the graph (i.e. y = x 2 ) to determine the ^-coordinate of 
the turning point: 

/(0) = (0) 2 = 

This corresponds to the point that we have previously calculated. 



92 



CHAPTER 7. DIFFERENTIAL CALCULUS 7.5 



Example 10: Calculation of Turning Points 



QUESTION 



Calculate the turning points of the graph of the function 



f(x) = 2x 3 - 9x 2 + 12x - 15 



SOLUTION 



Step I : Determine the derivative of f(x) 

Using the rules of differentiation we get: 

f'(x) = fix 2 - 18x+ 12 
Step 2 : Set f'(x) = and calculate x-coordinate of turning point 



6x - 18x + 12 


= 





x 2 - 3x + 2 


= 





(x-2)(x-l) 


= 






Therefore, the turning points are at x = 2 and x = 1. 
Step 3 : Substitute x-coordinate of turning point into f(x) to determine y-coordinates 

/(2) = 2(2) 3 -9(2) 2 + 12(2) -15 
= 16-36 + 24-15 
= -11. 



/(I) = 2(1) J -9(1) 2 + 12(1)-15 
= 2-9 + 12-15 
= -10 

Step 4 : Write final answer 

The turning points of the graph of /(x) = 2x 3 — 9x 2 + 12x — 15 are (2; —11) 
and (1;-10). 



We are now ready to sketch graphs of functions. 

Method: 

Sketching Graphs: Suppose we are given that f(x) = ax 3 + bx 2 + ex + d, then there are five steps to 
be followed to sketch the graph of the function: 



1 . Determine the value of the ?/-intercept by substituting x = into f(x) 



93 



7.5 CHAPTER 7. DIFFERENTIAL CALCULUS 



2. Determine the x-intercepts by factorising ax 3 + bx 2 + ex + d = and solving for x. First try 
to eliminate constant common factors, and to group like terms together so that the expression is 
expressed as economically as possible. Use the factor theorem if necessary. 

3. Find the turning points of the function by working out the derivative ^ and setting it to zero, 
and solving for x. 

4. Determine the j/-coordinates of the turning points by substituting the x values obtained in the 
previous step, into the expression for f(x). 

5. Use the information you're given to plot the points and get a rough idea of the gradients between 
points. Then fill in the missing parts of the function in a smooth, continuous curve. 



Example 11: Sketching Graphs 



QUESTION 



Draw the graph of g(x) = x 2 — x + 2 



SOLUTION 



Step 7 : Determine the shape of the graph 

The leading coefficient of x is > therefore the graph is a parabola with a 
minimum. 

5tep 2 : Determine the y-intercept 

The ^-intercept is obtained by setting x = 0. 

9 (0) = (0) 2 -0 + 2 = 2 
The turning point is at (0: 2). 

Step 3 : Determine the x-intercepts 

The x-intercepts are found by setting g{x) = 0. 

g(x) = x — x + 2 
= x 2 - x + 2 

Using the quadratic formula and looking at b 2 — iac we can see that this would 
be negative and so this function does not have real roots. Therefore, the graph of 
g(x) does not have any x-intercepts. 

Step 4 : Find the turning points of the function 

Work out the derivative ^ and set it to zero to for the x coordinate of the turning 
point. 

^=2x-l 
ax 







dx 
2x- 1 

2x = 1 
1 

X = 2 



94 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.5 



Step 5 : Determine the y-coordinates of the turning points by substituting the x values 
obtained in the previous step, into the expression for f(x). 

y coordinate of turning point is given by calculating g{\). 



V 

1 1 

4 ~ 2 

7 

4 



The turning point is at (|; J) 



Step 6 : Draw a neat sketch 




-3 -2 -1 



I 1 1 h» * 

12 3 4 



Example 12: Sketching Graphs 



QUESTION 



Sketch the graph of g(x) = -x 3 + 6x 2 - 9x + 4. 



SOLUTION 



Step 1 : Determine the shape of the graph 

The leading coefficient of x is < therefore the graph is a parabola with a 
maximum. 



95 



7.5 CHAPTER 7. DIFFERENTIAL CALCULUS 



Step 2 : Determine the y-intercept 

We find the y-intercepts by finding the value for 9(0). 

g(x) = — x + 6x — 9x + 4 
y int = g(0) = -(0) 3 +6(0) 2 -9(0)+4 
= 4 

Step 3 : Determine the x-intercepts 

We find the x-intercepts by finding the points for which the function g(x) = 0. 

g(x) = —x + 6x — 9x + 4 

Use the factor theorem to confirm that (x — 1) is a factor. If g(l) = 0, then 
(x — 1) is a factor. 

g(x) = —x + 6x — 9x + 4 

9(1) = -(l) 3 +6(l) 2 -9(l) + 4 

= -1 + 6-9 + 4 

= 

Therefore, (x — 1) is a factor. 

If we divide g(x) by (x — 1) we are left with: 

—x + 5x — 4 

This has factors 

-(x-4)(x- 1) 

Therefore: 

9 (x) = -(x-l)(x-l)(x-4) 
The x-intercepts are: x int = 1; 4 

Step 4 : Calculate the turning points 

Find the turning points by setting g'(x) = 0. 
If we use the rules of differentiation we get 

g'(x) = -3x 2 + 12x-9 

g'(x) = 

-3x 2 + 12x - 9 = 

x 2 - 4x + 3 = 

(x-3)(x-l) = 

clearpage The x-coordinates of the turning points are: x = 1 and x = 3. 
The ^/-coordinates of the turning points are calculated as: 



9{x) 


= 


—x + 6x — 9x 


+ 4 


9(1) 


= 


-(1) 3 +6(1) 2 - 


9(1) + 4 




= 


-1+6-9+4 






= 








96 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.5 



g(x) = — x + 6x — 9x + 4 

3(3) = -(3) 3 +6(3) 2 -9(3) + 4 

= -27 + 54-27 + 4 

= 4 

Therefore the turning points are: (1; 0) and (3; 4). 



Step 5 : Draw a neat sketch 




Exercise 7-4 



1 . Given f(x) = x 3 + x 2 - 5x + 3: 

(a) Show that (x — 1) is a factor of f(x) and hence factorise f(x) fully. 

(b) Find the coordinates of the intercepts with the axes and the turning points and sketch the 
graph 

2. Sketch the graph of f(x) = x :i — Ax 1 — llx + 30 showing all the relative turning points and 
intercepts with the axes. 

3. (a) Sketch the graph of f(x) = x 3 — 9x 2 + 24x — 20, showing all intercepts with the axes and 

turning points. 

(b) Find the equation of the tangent to f(x) at x = 4. 



97 



7.5 



CHAPTER 7. DIFFERENTIAL CALCULUS 



l\n More practice Cr) video solutions f'fj or help at www.everythingmaths.co.za 



(1.)01fy (2.)01fz (3.)01gO 



Local Minimum, Local Maximum and Point 
of Inflection 



EMCBM 



If the derivative (J 2 ) is zero at a point, the gradient of the tangent at that point is zero. It means that a 
turning point occurs as seen in the previous example. 




From the drawing the point (1; 0) represents a local minimum and the point (3; 4) the local maximum. 

A graph has a horizontal point of inflexion where the derivative is zero but the sign of the gradient 
does not change. That means the graph will continue to increase or decrease after the stationary point. 



MS 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.6 




From this drawing, the point (3; 1) is a horizontal point of inflexion, because the sign of the derivative 
does not change from positive to negative. 




7.6 Using Differential Calculus to 
Solve Problems 



We have seen that differential calculus can be used to determine the stationary points of functions, in 
order to sketch their graphs. However, determining stationary points also lends itself to the solution of 
problems that require some variable to be optimised. 

For example, if fuel used by a car is defined by: 



^-v 2 -6v + 245 

oU 



(7.19) 



where v is the travelling speed, what is the most economical speed (that means the speed that uses the 
least fuel)? 

If we draw the graph of this function we find that the graph has a minimum. The speed at the minimum 
would then give the most economical speed. 



60 

§ 50 

c 
O 
E. 40 

E 

| 30 

o 
u 

1 20 
10 




o + — I 1 1 1 1 1 1 1 1 1 1 1 1 \— 

10 20 30 40 50 60 70 80 90 100 110 120 130 140 

speed (km/h) 



99 



7.6 CHAPTER 7. DIFFERENTIAL CALCULUS 



We have seen that the coordinates of the turning point can be calculated by differentiating the function 
and finding the x-coordinate (speed in the case of the example) for which the derivative is 0. 

Differentiating (7.19), we get: 

f'(v) = -v-6 

If we set f'(v) = we can calculate the speed that corresponds to the turning point. 



° = l V - 6 
6 x 40 

V = -3~ 
= 80 



This means that the most economical speed is 80 km/h. 
See video: VMhgi at www.everythingmaths.co.za 



Example 13: Optimisation Problems 



QUESTION 


The sum of two positive numbers is 10. One of the numbers is multiplied by the square of the 
other. If each number is greater than 0, find the numbers that make this product a maximum. 


SOLUTION 




Step 1 


Examine the problem and formulate the equations that are required 

Let the two numbers be a and b. Then we have: 






a + b = 10 


(7.20) 




We are required to minimise the product of a and b. Call the proc 
Then: 


uct P. 




P = a.b 


(7.21) 




We can solve for b from (7.20) to get: 






b = 10 - a 


(7.22) 




Substitute this into (7.21) to write P in terms of a only. 






P = a (10 - a) = 10a - a 2 


(7.23) 


Step 2 


Differentiate 

The derivative of (7.23) is: 

P'(a) = 10 - 2a 




Step 3 


Find the stationary point 

Set P'{a) = to find the value of a which makes P a maximum. 





100 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.6 



P'(a) 


= 


10- 


-2a 





= 


10- 


-2a 


2a 


= 


10 




a 


= 


10 
2 




a 


= 


5 





Substitute into (7.26) to solve for the width. 

b = 10 - a 
= 10-5 
= 5 

Step 4 : Write the final answer 

The product is maximised when a and b are both equal to 5. 



Example 14: Optimisation Problems 



QUESTION 



Michael wants to start a vegetable garden, which he decides to fence off in the shape of a 
rectangle from the rest of the garden. Michael has only 160 m of fencing, so he decides to use 
a wall as one border of the vegetable garden. Calculate the width and length of the garden 
that corresponds to largest possible area that Michael can fence off. 




width, w 



SOLUTION 



Step I : Examine the problem and formulate the equations that are required 

The important pieces of information given are related to the area and modified 
perimeter of the garden. We know that the area of the garden is: 



.1 



(7.24) 



We are also told that the fence covers only 3 sides and the three sides should add 
up to 160 m. This can be written as: 



160 m = w + 1 + 1 



(7.25) 



101 



7.6 CHAPTER 7. DIFFERENTIAL CALCULUS 

However, we can use (7.25) to write w in terms of /: 

w = 160m -21 (7.26) 

Substitute (7.26) into (7.24) to get: 

A = (160 m - 21)1 = 160 m - 21 2 (7.27) 

Step 2 : Differentiate 

Since we are interested in maximising the area, we differentiate (7.27) to get: 

A' (I) = 160 m -41 

Step 3 : Find the stationary point 

To find the stationary point, we set A'(l) = and solve for the value of I that 
maximises the area. 

A' (I) = 160 m -41 

= 160m -41 

:Al = 160m 
160 m 



4 
I = 40 m 

Substitute into (7.26) to solve for the width. 

w = 160 m -21 

= 160 m- 2(40 m) 

= 160 m -80 m 

= 80 m 

Step 4 : Write the final answer 

A width of 80 m and a length of 40 m will yield the maximal area fenced off. 



Exercise 7-5 



1 . The sum of two positive numbers is 20. One of the numbers is multiplied by the square of the 
other. Find the numbers that make this product a maximum. 



2. A wooden block is made as shown in the diagram. The ends are right-angled triangles having 
sides 3x, Ax and 5x. The length of the block is y. The total surface area of the block is 3600 cm 2 . 



102 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.6 




(a) Show that y 



1200cm 2 - 4x 2 

ox 



(b) Find the value of x for which the block will have a maximum volume. (Volume = area of 
base x height.) 

3. The diagram shows the plan for a veranda which is to be built on the corner of a cottage. A 
railing ABCDE is to be constructed around the four edges of the veranda. 



c 




y 






D 


J 




L 






verandah 




X 




1 


F 


_C 




B 


r 


J 
A 


E 


















cottage 



If AB = DE = x and BC = CD = y, and the length of the railing must be 30 m, find the 
values of x and y for which the verandah will have a maximum area. 



More practice f ►) video solutions CfJ or help at www.everythingmaths.co.za 



(1.)01g4 (2.)01g5 (3.) 01; 



Rate of Change Problems 



EMCBO 



Two concepts were discussed in this chapter: Average rate of change = f( - b l _^ (a) and Instantaneous 



rate of change = lim^o 



f(x + h)-.f(x) 



When we mention rate of change, the latter is implied. Instan- 



11).', 



7.6 CHAPTER 7. DIFFERENTIAL CALCULUS 



taneous rate of change is the derivative. When average rate of change is required, it will be specifically 
referred to as average rate of change. 

Velocity is one of the most common forms of rate of change. Again, average velocity = average rate 
of change and instantaneous velocity = instantaneous rate of change = derivative. Velocity refers to 
the increase of distance(s) for a corresponding increase in time (t). The notation commonly used for 
this is: 

v(t) = J = ,'(*) 

where s'(t) is the position function. Acceleration is the change in velocity for a corresponding increase 
in time. Therefore, acceleration is the derivative of velocity 

a(t) = v'(t) 

This implies that acceleration is the second derivative of the distance(s). 



Example 15: Rate of Change 



QUESTION 



The height (in metres) of a golf ball that is hit into the air after t seconds, is given by h(t) 
20* - 5t 2 . Determine 



1 . the average velocity of the ball during the first two seconds 

2. the velocity of the ball after 1,5 s 

3. the time at which the velocity is zero 

4. the velocity at which the ball hits the ground 

5. the acceleration of the ball 

SOLUTION 



Step 1 : Average velocity 



= 10m-s 

Step 2 : Instantaneous Velocity 

v(t) 



h{2) - /t(0) 

2-0 
[20(2) - 5(2) 2 ] - [20(0) - 5(0) 2 

2 
40-20 

2 

-l 



dh 
dt 
20 - 10* 



104 



CHAPTER 7. DIFFERENTIAL CALCULUS 



7.6 



Velocity after 1,5 s: 



v(l,5) = 20-10(1,5) 

r -1 

= 5m ■ s 



Step 3 : Zero velocity 





v(t) 


= 





20 


- 104 


= 







Wt 


= 


20 




t 


= 


2 



Therefore the velocity is zero after 2 s 

Step 4 : Ground velocity 

The ball hits the ground when hit) = 

20* - 5i 2 = 
5t(4 - 1) = 
t = or t = 4 



The ball hits the ground after 4s. The velocity after 4s will be: 

1,(4) = h'{A) 

= 20 - 10(4) 
= -20 m-s -1 



The ball hits the ground at a speed of 20 m ■ s _1 . Notice that the sign of the 
velocity is negative which means that the ball is moving downward (the reverse 
of upward, which is when the velocity is positive). 



Step 5 : Acceleration 



a = v (t) 

= —10 m ■ s~ 

Just because gravity is constant does not mean we should necessarily think of 
acceleration as a constant. We should still consider it a function. 



Chapter 7 



End of Chapter Exercises 



1 . Determine f'(x) from first principles if: 



105 



7.6 CHAPTER 7. DIFFERENTIAL CALCULUS 



3. Determine ^f- if: 

dy 



2. Given: f(x) 
Dete 

(a) 

(b) 







f(x) = x - 


- (ix 








f(x) = 2x - 


2 
- X 




3.x, 


find 


f'(x) using first print 


ziples 






y = (2x) 2 - 


1 

3.x 








2V5- 


-5 





\/x 
2 



4. Given: f(x) = x 3 - 3x 2 + 4 

(a) Calculate /(— 1), and hence solve the equation /(x) = 

(b) Determine f'(x) 

(c) Sketch the graph of / neatly and clearly, showing the co-ordinates of the turning 
points as well as the intercepts on both axes. 

(d) Determine the co-ordinates of the points on the graph of / where the gradient is 
9. 

5. Given: /(x) = 2x 3 — 5x 2 — 4x + 3. The x-intercepts of / are: (— 1;0) (|;0) and 
(3;0). 

(a) Determine the co-ordinates of the turning points of /. 

(b) Draw a neat sketch graph of /. Clearly indicate the co-ordinates of the intercepts 
with the axes, as well as the co-ordinates of the turning points. 

(c) For which values of k will the equation /(x) = k , have exactly two real roots? 

(d) Determine the equation of the tangent to the graph of f(x) = 2x 3 — 5x 2 — 4x + 3 
at the point where x = 1. 

6. (a) Sketch the graph of /(x) = x 3 — 9x 2 + 24x — 20, showing all intercepts with the 

axes and turning points, 
(b) Find the equation of the tangent to /(x) at x = 4. 

7. Calculate: 

1-x 3 

lim 

x-n 1 — x 

8. Given: 

fix) = 2x — x 

(a) Use the definition of the derivative to calculate f'(x). 

(b) Hence, calculate the co-ordinates of the point at which the gradient of the tan- 
gent to the graph of / is 7. 

9. If xy — 5 = vV, determine j^ 

10. Given: g(x) = (x~ 2 +x 2 ) 2 . Calculate g' (2). 

11. Given: /(x) = 2x - 3 

(a) Find: / _1 (x) 

(b) Solve: / _1 (x) = 3/'(x) 

12. Find f'(x) for each of the following: 



(a) fix) = -^- + 10 

(b) f(x) = (2x»-5)(3x + 2) 

x 2 

1 3. Determine the minimum value of the sum of a positive number and its reciprocal. 

14. If the displacement s (in metres) of a particle at time t (in seconds) is governed by the 
equation s = \t 3 — 2t, find its acceleration after 2 seconds. (Acceleration is the rate 
of change of velocity, and velocity is the rate of change of displacement.) 



106 



CHAPTER 7. DIFFERENTIAL CALCULUS 7.6 



1 5. (a) After doing some research, a transport company has determined that the rate at 
which petrol is consumed by one of its large carriers, travelling at an average 
speed of x km per hour, is given by: 



55 a- 1 
P(x) ■■ 



2x 200 km 2 ^- 1 !!- 1 

i. Assume that the petrol costs R4,00 per litre and the driver earns R18,00 per 
hour (travelling time). Now deduce that the total cost, C, in Rands, for a 
2 000 km trip is given by: 

_, , 256000 kmh _1 R ,„ 

C(x) = h 40x 

x 

ii. Hence determine the average speed to be maintained to effect a minimum 
cost for a 2 000 km trip. 
(b) During an experiment the temperature T (in degrees Celsius), varies with time t 
(in hours), according to the formula: 

T{t) = 30 + 4t- i* 2 te[l;10] 

i. Determine an expression for the rate of change of temperature with time, 
ii. During which time interval was the temperature dropping? 

16. The depth, d, of water in a kettle t minutes after it starts to boil, is given by d = 
86 — it — jt\ where d is measured in millimetres. 

(a) How many millimetres of water are there in the kettle just before it starts to boil? 

(b) As the water boils, the level in the kettle drops. Find the rate at which the water 
level is decreasing when t = 2 minutes. 

(c) How many minutes after the kettle starts boiling will the water level be dropping 
at a rate of 12| mm/minute? 



(ft" 1 ) More practice (►) video solutions CfJ or help at www.everythingmaths.co.za 

(1.)01g7 (2.)01g8 (3.)01g9 (4.)01ga (5.) 01gb (6.) Olgc 
(7.)01gd (8.)01ge (9.)01gf (10.) 01gg (11.)01gh (12.) 01 gi 
(13.)01gj (14.)01gk (15.)01gm (16.)021h 



107 



Linear Programming 





8. 1 Introduction 




In Grade 11 you were introduced to linear programming and solved problems by looking at points 
on the edges of the feasible region. In Grade 12 you will look at how to solve linear programming 
problems in a more general manner. 

© See introductory video: VMhgw at www.everythingmaths.co.za 




8.2 Terminology 




Feasible Region and Points 



EMCBR 



Tip 



The constraints are used 
to create bounds of the 
solution. 



Tip 



ax + by = c 

o If b ^ 0, feasible 
points must lie on the 






o If b = 0, feasible 
points must lie on the 
line x = c/a 

ax + by < c 

o If b jL 0, feasible 
points must lie on 
or below the line 

o If b = 0, feasible 
points must lie on or 
to the left of the line 
x = c/a. 



Constraints mean that we cannot just take any x and y when looking for the x and y that optimise our 
objective function. If we think of the variables x and y as a point (x; y) in the xj/-plane then we call 
the set of all points in the xj/-plane that satisfy our constraints the feasible region. Any point in the 
feasible region is called a feasible point. 



For example, the constraints 



x > 
y>0 



mean that every (x,y) we can consider must lie in the first quadrant of the xy plane. The constraint 

x>y 
means that every (x,y) must lie on or below the line y = x and the constraint 

x < 20 

means that x must lie on or to the left of the line x = 20. 

We can use these constraints to draw the feasible region as shown by the shaded region in Figure 8.1 . 

When a constraint is linear, it means that it requires that any feasible point (x,y) lies on one side of 
or on a line. Interpreting constraints as graphs in the xy plane is very important since it allows us to 
construct the feasible region such as in Figure 8.1. 



108 



CHAPTER 8. LINEAR PROGRAMMING 



8.3 




5 10 15 20 
Figure 8.1: The feasible region corresponding to the constraints x >0, y >0, x > y and x < 20. 




8.3 Linear Programming and the 
Feasible Region 




If the objective function and all of the constraints are linear then we call the problem of optimising 
the objective function subject to these constraints a linear program. All optimisation problems we will 
look at will be linear programs. 

The major consequence of the constraints being linear is that the feasible region is always a polygon. 
This is evident since the constraints that define the feasible region all contribute a line segment to its 
boundary (see Figure 8.1). It is also always true that the feasible region is a convex polygon. 

The objective function being linear means that the feasible point(s) that gives the solution of a linear 
program always lies on one of the vertices of the feasible region. This is very important since, as we 
will soon see, it gives us a way of solving linear programs. 

We will now see why the solutions of a linear program always lie on the boundary of the feasible 
region. Firstly, note that if we think of f(x,y) as lying on the z axis, then the function f(x,y) = ax + by 
(where a and b are real numbers) is the definition of a plane. If we solve for y in the equation defining 
the objective function then 



f(x,y) 



ax + by 

f(x,y) 



(8.1) 



What this means is that if we find all the points where f(x,y) = c for any real number c (i.e. f(x,y) 
is constant with a value of c), then we have the equation of a line. This line we call a level line of the 
objective function. 

Consider again the feasible region described in Figure 8.1. Let's say that we have the objective function 
f(x,y) = x — 2y with this feasible region. If we consider Equation 8.3 corresponding to 



then we get the level line 



f(x,y) = -20 



y = -x + 10 
y 2 



10!) 



8.3 



CHAPTER 8. LINEAR PROGRAMMING 



which has been drawn in Figure 8.2. Level lines corresponding to 

x 



f(x,y) = -10 


or 


y=2 + 5 


/(*,!/) = 


or 


X 

V= 2 


f(x,y) = 10 


or 


X K 

y = x — 5 

y 2 


f(x,y) = 20 


or 


»=f-io 



have also been drawn in. It is very important to realise that these are not the only level lines; in fact, 
there are infinitely many of them and they are all parallel to each other. Remember that if we look at 
any one level line f(x,y) has the same value for every point (x,y) that lies on that line. Also, f(x,y) 
will always have different values on different level lines. 




10 AS 



Figure 8.2: The feasible region corresponding to the constraints x > 0, y > 0, x > y and x < 20 with 
objective function f(x,y) = x — 2y, The dashed lines represent various level lines of f(x,y). 



If a ruler is placed on the level line corresponding to f(x,y) = —20 in Figure 8.2 and moved down the 
page parallel to this line then it is clear that the ruler will be moving over level lines which correspond 
to larger values of f(x,y). So if we wanted to maximise f(x,y) then we simply move the ruler down the 
page until we reach the bottom-most point in the feasible region. This point will then be the feasible 
point that maximises f(x,y). Similarly, if we wanted to minimise f(x,y) then the top-most feasible 
point will give the minimum value of f(x,y). 

Since our feasible region is a polygon, these points will always lie on vertices in the feasible region. 
The fact that the value of our objective function along the line of the ruler increases as we move it 
down and decreases as we move it up depends on this particular example. Some other examples 
might have that the function increases as we move the ruler up and decreases as we move it down. 

It is a general property, though, of linear objective functions that they will consistently increase or 
decrease as we move the ruler up or down. Knowing which direction to move the ruler in order to 
maximise/minimise f(x,y) = ax + by is as simple as looking at the sign of 6 (i.e. "is b negative, positive 
or zero?"). If b is positive, then f(x,y) increases as we move the ruler up and f(x,y) decreases as we 
move the ruler down. The opposite happens for the case when b is negative: f(x,y) decreases as we 
move the ruler up and f(x,y) increases as we move the ruler down. If b = 0, we need to look at the 
sign of a. 

If a is positive then f(x,y) increases as we move the ruler to the right and decreases if we move the 
ruler to the left. Once again, the opposite happens for a negative. If we look again at the objective 
function mentioned earlier, 

f(x,y) = x - 2y 

with a = 1 and b = —2, then we should find that f(x,y) increases as we move the ruler down the 
page since b = — 2 < 0. This is exactly what happened in Figure 8.2. 

The main points about linear programming we have encountered so far are 



110 



CHAPTER 8. LINEAR PROGRAMMING 8.3 



• The feasible region is always a polygon. 

• Solutions occur at vertices of the feasible region. 

• Moving a ruler parallel to the level lines of the objective function up/down to the top/bottom of 
the feasible region shows us which of the vertices is the solution. 

• The direction in which to move the ruler is determined by the sign of b and also possibly by the 
sign of a. 

These points are sufficient to determine a method for solving any linear program. 

Method: Linear Programming 

If we wish to maximise the objective function f(x,y) then: 

1 . Find the gradient of the level lines of f(x,y) (this is always going to be — | as we saw in Equa- 
tion ??) 

2. Place your ruler on the xy plane, making a line with gradient — | (i.e. b units on the x-axis and 
—a units on the y-axis) 

3. The solution of the linear program is given by appropriately moving the ruler. Firstly we need to 
check whether b is negative, positive or zero. 

(a) If b > 0, move the ruler up the page, keeping the ruler parallel to the level lines all the time, 
until it touches the "highest" point in the feasible region. This point is then the solution. 

(b) If b < 0, move the ruler in the opposite direction to get the solution at the "lowest" point in 
the feasible region. 

(c) If b = 0, check the sign of a 

i. If a < move the ruler to the "leftmost" feasible point. This point is then the solution, 
ii. If a > move the ruler to the "rightmost" feasible point. This point is then the solution. 



Example 1: Prizes! 



QUESTION 

As part of their opening specials, a furniture store promised to give away at least 40 prizes 
with a total value of at least R2 000. The prizes are kettles and toasters. 

1 . If the company decides that there will be at least 10 of each prize, write down two more 
inequalities from these constraints. 

2. If the cost of manufacturing a kettle is R60 and a toaster is R50, write down an objective 
function C which can be used to determine the cost to the company of both kettles and 
toasters. 

3. Sketch the graph of the feasibility region that can be used to determine all the possible 
combinations of kettles and toasters that honour the promises of the company. 

4. How many of each prize will represent the cheapest option for the company? 

5. How much will this combination of kettles and toasters cost? 



SOLUTION 



111 



8.3 



CHAPTER 8. LINEAR PROGRAMMING 



Step 1 : Identify the decision variables 

Let the number of kettles be x and the number of toasters be y and write down 
two constraints apart from x > and y > that must be adhered to. 

Step 2 : Write constraint equations 

Since there will be at least 10 of each prize we can write: 



x > 10 



and 



y> 10 
Also, the store promised to give away at least 40 prizes in total. Therefore: 

x + y > 40 

Step 3 : Write the objective function 

The cost of manufacturing a kettle is R60 and a toaster is R50. Therefore the cost 
the total cost C is: 

C = 60x + 50j/ 

Step 4 : Sketch the graph of the feasible region 





i 




















90 - 






















80 - 






















70 - 






















60 - 




















50 - 






















40 - 






















30 - 




B 


















20 - 






\ A 




















\ 




1 


1 

20 


i 
30 


40 


i 
50 


60 


70 


80 


90 


100 



Step 5 : Determine vertices of feasible region 

From the graph, the coordinates of vertex A are (30; 10) and the coordinates of 
vertex Bare (10; 30). 

Step 6 : Draw in the search line 

The search line is the gradient of the objective function. That is, if the equation 
C = dOx + 50y is now written in the standard form y = . . ., then the gradient is: 

6 
m = --, 

which is shown with the broken line on the graph. 



112 



CHAPTER 8. LINEAR PROGRAMMING 



8.3 




10 20 30 40 50 60 70 80 90 100 



Step 7 : Calculate cost at each vertex 

At vertex A, the cost is: 



C = 60x + 50|/ 

= 60(30) + 50(10) 

= 1 800 + 500 

= R2 300 

At vertex B, the cost is: 

C = 60x fc + 50»/t 

= 60(10) + 50(30) 

= 600 + 1500 

= R2 100 

Step 8 : Write the final answer 

The cheapest combination of prizes is 10 kettles and 30 toasters, costing the 
company R2 100. 



Example 2: Search Line Method 



QUESTION 



As a production planner at a factory manufacturing lawn cutters your job will be to advise 
the management on how many of each model should be produced per week in order to 
maximise the profit on the local production. The factory is producing two types of lawn 
cutters: Quadrant and Pentagon. Two of the production processes that the lawn cutters must 



113 



8.3 



CHAPTER 8. LINEAR PROGRAMMING 



go through are: bodywork and engine work. 

• The factory cannot operate for less than 360 hours on engine work for the lawn cutters. 

• The factory has a maximum capacity of 480 hours for bodywork for the lawn cutters. 

• Half an hour of engine work and half an hour of bodywork is required to produce one 
Quadrant. 

• The ratio of Pentagon lawn cutters to Quadrant lawn cutters produced per week must be 
at least 3 : 2. 

• A minimum of 200 Quadrant lawn cutters must be produced per week. 

Let the number of Quadrant lawn cutters manufactured in a week be x. 
Let the number of Pentagon lawn cutters manufactured in a week be y. 
Two of the constraints are: 

x > 200 

3x + 2y > 2 160 

7. Write down the remaining constrain ts in terms of x and y to represent the above men- 
tioned information. 

2. Use graph paper to represent the constraints graphically. 

3. Clearly indicate the feasible region by shading it. 

4. If the profit on one Quadrant lawn cutter is Rl 200 and the profit on one Pentagon lawn 
cutter is R400, write down an equation that will represent the profit on the lawn cutters. 

5. Using a search line and your graph, determine the number of Quadrant and Pentagon 
lawn cutters that will yield a maximum profit. 

6. Determine the maximum profit per week. 



SOLUTION 



Step I : Remaining constraints: 



1 



"-x+ -y <480 



i>± 



Step 2 : Graphical representation 




200 



720 960 



111 



CHAPTER 8. LINEAR PROGRAMMING 8.3 



Step 3 : Profit equation 



P = 1 200x + 400|/ 



Step 4 : Maximum profit 

By moving the search line upwards, we see that the point of maximum profit is 
at (600; 900). Therefore 

P = 1 200(600) + 400(900) 

p = m 080 ooo 



Chapter 8 



End of Chapter Exercises 



1. Polkadots is a small company that makes two types of cards, type X and type Y. 
With the available labour and material, the company can make not more than 150 
cards of type X and not more than 120 cards of type Y per week. Altogether they 
cannot make more than 200 cards per week. 

There is an order for at least 40 type X cards and 10 type Y cards per week. Polkadots 
makes a profit of R5 for each type X card sold and R10 for each type Y card. 
Let the number of type X cards be x and the number of type Y cards be y, manufac- 
tured per week. 

(a) One of the constraint inequalities which represents the restrictions above is x < 
150. Write the other constraint inequalities. 

(b) Represent the constraints graphically and shade the feasible region. 

(c) Write the equation that represents the profit P (the objective function), in terms 
of x and y. 

(d) On your graph, draw a straight line which will help you to determine how many 
of each type must be made weekly to produce the maximum P 

(e) Calculate the maximum weekly profit. 

2. A brickworks produces "face bricks" and "clinkers". Both types of bricks are pro- 
duced and sold in batches of a thousand. Face bricks are sold at R150 per thousand, 
and clinkers at R100 per thousand, where an income of at least R9 000 per month is 
required to cover costs. The brickworks is able to produce at most 40 000 face bricks 
and 90 000 clinkers per month, and has transport facilities to deliver at most 100 000 
bricks per month. The number of clinkers produced must be at least the same as the 
number of face bricks produced. 

Let the number of face bricks in thousands be x, and the number of clinkers in 
thousands be y, 

(a) List all the constraints. 

(b) Graph the feasible region. 

(c) If the sale of face bricks yields a profit of R25 per thousand and clinkers R45 per 
thousand, use your graph to determine the maximum profit. 

(d) If the profit margins on face bricks and clinkers are interchanged, use your graph 
to determine the maximum profit. 



115 



8.3 CHAPTER 8. LINEAR PROGRAMMING 



3. A small cell phone company makes two types of cell phones: Easyhear and Longtalk. 
Production figures are checked weekly. At most, 42 Easyhear and 60 Longtalk phones 
can be manufactured each week. At least 30 cell phones must be produced each 
week to cover costs. In order not to flood the market, the number of Easyhear phones 
cannot be more than twice the number of Longtalk phones. It takes | hour to as- 
semble an Easyhear phone and | hour to put together a Longtalk phone. The trade 
unions only allow for a 50-hour week. 

Let x be the number of Easyhear phones and y be the number of Longtalk phones 
manufactured each week. 

(a) Two of the constraints are: 

< x < 42 and < y < 60 

Write down the other three constraints. 

(b) Draw a graph to represent the feasible region 

(c) If the profit on an Easyhear phone is R225 and the profit on a Longtalk is R75, 
determine the maximum profit per week. 

4. Hair for Africa is a firm that specialises in making two kinds of up-market shampoo, 
Glowhair and Longcurls. They must produce at least two cases of Glowhair and one 
case of Longcurls per day to stay in the market. Due to a limited supply of chemicals, 
they cannot produce more than 8 cases of Glowhair and 6 cases of Longcurls per 
day. It takes half-an-hour to produce one case of Glowhair and one hour to produce 
a case of Longcurls, and due to restrictions by the unions, the plant may operate for 
at most 7 hours per day. The workforce at Hair for Africa, which is still in training, 
can only produce a maximum of 10 cases of shampoo per day. 

Let x be the number of cases of Glowhair and y the number of cases of Longcurls 
produced per day. 

(a) Write down the inequalities that represent all the constraints. 

(b) Sketch the feasible region. 

(c) If the profit on a case of Glowhair is R400 and the profit on a case of Longcurls 
is R300, determine the maximum profit that Hair for Africa can make per day. 

5. A transport contractor has six 5-ton trucks and eight 3-ton trucks. He must deliver 
at least 120 tons of sand per day to a construction site, but he may not deliver more 
than 180 tons per day. The 5-ton trucks can each make three trips per day at a cost of 
R30 per trip, and the 3-ton trucks can each make four trips per day at a cost of R120 
per trip. How must the contractor utilise his trucks so that he has minimum expense? 



More practice (►) video solutions C{J or help at www.everythingmaths.< 
(1.)0lgn (2.)01gp (3.)01gq (4.) Olgr (5.)01gs 



116 



Geometry 





9.1 Introduction 




In previous years, you learned about the geometry of points, lines and various polygons, made up from 
lines. Here we will discuss the geometry of circle in a lot of depth. 

© See introductory video: VMhmc at www.everythingmaths.co.za 




9.2 Circle Geometry 




Terminology 



EMCBV 



The following is a recap of terms that are regularly used when referring to circles. 

arc An arc is a part of the circumference of a circle. 

chord A chord is a straight line joining the ends of an arc. 

radius A radius, r, is any straight line from the centre of the circle to a point on the circumference. 

diameter A diameter, 0, is a special chord that passes through the centre of the circle. A diameter is 
the length of a straight line segment from one point on the circumference to another point on 
the circumference, that passes through the centre of the circle. 

segment A segment is the part of the circle that is cut off by a chord. A chord divides a circle into two 
segments. 

tangent A tangent is a line that makes contact with a circle at one point on the circumference. (AB is 
a tangent to the circle at point P) in Figure 9.1. 



Axioms 



EMCBW 



An axiom is an established or accepted principle. For this section, the following are accepted as 
axioms. 



117 



9.2 



CHAPTER 9. GEOMETRY 




p tangent 
Figure 9.1 : Parts of a circle 




Figure 9.2: A right-angled triangle 



1. The Theorem of Pythagoras, which states that the square on the hypotenuse of a right-angled 
triangle is equal to the sum of the squares on the other two sides. In AABC, this means that 

{ABf + (BC) 2 = (AC) 2 



2. A tangent is perpendicular to the radius, drawn at the point of contact with the circle. 



Theorems of the Geometry of Circles 



EMCBX 



A theorem is a general proposition that is not self-evident but is proved by reasoning (these proofs need 
not be learned for examination purposes). 



Theorem 1. The line drawn from the centre of a circle, perpendicular to a chord, bisects the chord. 



118 



CHAPTER 9. GEOMETRY 



9.2 



Proof: 




Consider a circle, with centre O. Draw a chord AB and draw a perpendicular line from the centre of 
the circle to intersect the chord at point P. 

The aim is to prove that AP = BP 

1 . AOAP and AOBP are right-angled triangles. 

2. OA = OB as both of these are radii and OP is common to both triangles. 

Apply the Theorem of Pythagoras to each triangle, to get: 

OA 2 = OP 2 + AP 2 



OB 2 



OP 2 + BP 2 



However, OA = OB. So, 



OP + AP 2 

:. AP 2 

and AP 



OP + BP 2 

BP 2 

BP 



This means that OP bisects AB. 

Theorem 2. The line drawn from the centre of a circle, that bisects a chord, is perpendicular to the 
chord. 

Proof: 




Consider a circle, with centre O. Draw a chord AB and draw a line from the centre of the circle to 
bisect the chord at point P. 

The aim is to prove that OP ± AB 

In AOAP and AOBP, 



119 



9.2 



CHAPTER 9. GEOMETRY 



1 . AP = PB (given) 



2. OA = OB (radii 



3. OP is common to both triangles. 



AOAP = AOBP (SSS). 



OPA 

OP A + OPB 

:. OPA 

:.OP 



OPB 

180° (APB is a straight line) 

OPB = 90° 

AB 



Theorem 3. The perpendicular bisector of a chord passes through the centre of the circle. 



Proof: 




Consider a circle. Draw a chord AB. Draw a line PQ perpendicular to AB such that PQ bisects AB 
at point P. Draw lines AQ and BQ. 

The aim is to prove that Q is the centre of the circle, by showing that AQ = BQ. 

In AOAP and AOBP, 

1 . AP = PB (given) 

2. ZQP.4 = ZQPB (QP _L AB) 



3. QP is common to both triangles. 



.-. AQAP = AQBP (SAS). 

From this, QA = QB. Since the centre of a circle is the only point inside a circle that has points on 
the circumference at an equal distance from it, Q must be the centre of the circle. 



120 



CHAPTER 9. GEOMETRY 



9.2 



Exercise 9-1 



Find the value of x: 




2. 



5. 






f/Vy More practice ( ►) video solutions (*?) or help at www.everythingmaths.co.za 



(1.)021i (2.) 021j (3.) 021k (4.) 021m (5.) 021n (6.) 021p 



121 



9.2 



CHAPTER 9. GEOMETRY 



Theorem 4. The angle subtended by an arc at the centre of a circle is double the size of the angle 
subtended by the same arc at the circumference of the circle. 



Proof: 




Consider a circle, with centre O and with A and B on the circumference. Draw a chord AB. Draw 
radii OA and OB. Select any point P on the circumference of the circle. Draw lines PA and PB. 
Draw PO and extend to R. 

The aim is to prove that AOB = 2 . APB. 

AOR = PAO + APO (exterior angle = sum of interior opp. angles) 

But, PAO = APO {AAOP is an isosceles A) 

.-. AOR = 2APO 

Similarly, BOR = 2BPO. 

So, 



AOB 



AOR + BOR 
2APO + 2BPO 
2(APO + BPO) 
2(APB) 



122 



CHAPTER 9. GEOMETRY 



9.2 



Exercise 9-2 



Find the angles (a to /) indicated in each diagram: 






4. K 






fa*) More practice CrJ video solutions Qfj or help at www.everythingmaths.co.za 



(1.)021q (2.)021r (3.) 021s (4.) 021t (5.)021u (6.) 021v 

Theorem 5. The angles subtended by a chord at the circumference of a circle are equal, if the angles 
are on the same side of the chord. 



Proof: 




12:', 



9.2 



CHAPTER 9. GEOMETRY 



Consider a circle, with centre O. Draw a chord AB. Select any points P and Q on the circumference 
of the circle, such that both P and Q are on the same side of the chord. Draw lines PA, PB, QA and 
QB. 

The aim is to prove that AQB = APB. 



AOB -- 


= 2AQB (Z at centre 


and AOB = 


= 2APB (Z at centre 


.-. 2AQB -- 


= 2APB 


:. AQB -- 


= APB 



twice Z at circumference (Theorem 4)) 
twice Z at circumference (Theorem 4)) 



Theorem 6. (Converse of Theorem 5) If a line segment subtends equal angles at two other points on 
the same side of the line, then these four points lie on a circle. 



Proof: 




Consider a line segment AB, that subtends equal angles at points P and Q on the same side of AB. 

The aim is to prove that points A, B, P and Q lie on the circumference of a circle. 

By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let 
the circle cut AP (or AP extended) at point R. 



AQB 

but AQB 

.'. ARB 

but this cannot be true since ARB 



ARB(Zs on same side of chord (Theorem 5)) 

APB (given) 

APB 

APB + RBP (exterior Z of A) 



.•. the assumption that the circle does not pass through P, must be false, and A, B, P and Q lie on the 
circumference of a circle. 



Exercise 9-3 



Find the values of the unknown letters. 



12 1 



CHAPTER 9. GEOMETRY 



9.2 







Q\*j More practice CwJ video 



solutions 



C'f) or help at www.everythingmaths.co.za 



(1.)021w (2.)021x (3.)021y (4.) 021z (5.) 0220 (6.) 0221 

Cyclic Quadrilaterals 

Cyclic quadrilaterals are quadrilaterals with all four vertices lying on the circumference of a circle. The 
vertices of a cyclic quadrilateral are said to be concyclic. 

Theorem 7. The opposite angles of a cyclic quadrilateral are supplementary. 
Proof: 



125 



9.2 



CHAPTER 9. GEOMETRY 




Consider a circle, with centre O. Draw a cyclic quadrilateral ABPQ. Draw AO and PO. 
The aim is to prove that ABP + AQP = 180° and QAB + QPB = 180°. 



Oi 

o 2 

But, Oi+Os 

.-. 2ABP + 2AQP 

.-. ABP + AQP 

Similarly, QAB + QPB 



2ABP (Zs at centre (Theorem 4)) 

2AQP (Zs at centre (Theorem 4)) 

360° 

360° 

180° 

180° 



Theorem 8. (Converse of Theorem 7) If the opposite angles of a quadrilateral are supplementary, then 
the quadrilateral is cyclic. 

Proof: 




Consider a quadrilateral ABPQ, such that ABP + AQP = 180° and QAB + QPB = 180°. 

The aim is to prove that points A, B, P and Q lie on the circumference of a circle. 

By contradiction. Assume that point P does not lie on a circle drawn through points A, B and Q. Let 
the circle cut QP (or QP extended) at point R. Draw BR. 



QAB + QRB 

but QAB + QPB 

.-.QRB 

but this cannot be true since QRB 



180° (opp. Zs of cyclic quad. (Theorem 7)) 

180° (given) 

QPB 

QPB + RBP (exterior Z of A) 



120 



CHAPTER 9. GEOMETRY 



9.2 



,\ the assumption that the circle does not pass through P, must be false, and A, B, P and Q lie on the 
circumference of a circle and ABPQ is a cyclic quadrilateral. 



Exercise 9-4 



Find the values of the unknown letters. 






f/Vy More practice (►) video solutions (cj or help at www.everythingmaths.co.za 



(1.)0222 (2.) 0223 (3.) 0224 (4.) 0225 



Theorem 9. Two tangents drawn to a circle from the same point outside the circle are equal in length. 



Proof: 



127 



9.2 



CHAPTER 9. GEOMETRY 




Consider a circle, with centre O. Choose a point P outside the circle. Draw two tangents to the circle 
from point P, that meet the circle at A and B. Draw lines OA, OB and OP. 

The aim is to prove that AP = BP. 

In AOAP and AOBP, 



1. OA = OB (radii) 

2. AOAP = ZOBP = 90° (OA _L AP and OB 1 BP) 

3. OP is common to both triangles. 

AOAP = AOBP (right angle, hypotenuse, side) 
.-. AP = BP 



Exercise 9-5 



Find the value of the unknown lengths. 



128 



CHAPTER 9. GEOMETRY 



9.2 



AE = 5 cm 
AC = 8 cm 
CE = 9 cm 





4. 




LN = 7.5 cm 



More practice (►) video solutions Cf) or help at www.everythingmaths 



(1.)0226 (2.) 0227 (3.) 0228 (4.) 0229 

Theorem 10. The angle between a tangent and a chord, drawn at the point of contact of the chord, is 
equal to the angle which the chord subtends in the alternate segment. 

Proof: 




Consider a circle, with centre O. Draw a chord AB and a tangent SR to the circle at point B. Chord 
AB subtends angles at points P and Q on the minor and major arcs, respectively. 

Draw a diameter BT and join A to T. 

The aim is to prove that APB = ABR and AQB = ABS. 

First prove that AQB = ABS as this result is needed to prove that APB = ABR. 



120 



9.2 CHAPTER 9. GEOMETRY 



ABS + ABT = 90° (TB ± SR) 

BAT = 90° (Zs at centre) 

.-. ABT + ATB = 90° (sum of angles in ABAT) 

:. ABS = ATB 

However, AQB = ATB (angles subtended by same chord AB (Theorem 5)) 

.-. AQB = ABS 

ABS + ABR = 180° (SBR is a straight line) 

APB + AQB = 180° (APBQ is a cyclic quad. (Theorem 7) 

From (9.1), AQB = ABS 

.-. 180° - AQB = 180° - ABS 

.-. APB = ABR 

(9.1) 



Exercise 9-6 



Find the values of the unknown letters. 



i;so 



CHAPTER 9. GEOMETRY 



9.2 




3. O 






131 



9.2 



CHAPTER 9. GEOMETRY 



More practice ( ►) video solutions f?j or help at www.everythingmaths.co.za 



(1.)022a (2.) 022b (3.) 022c (4.) 022d (5.) 022e (6.) 022f 
(7.) 022g (8.) 022h 

Theorem 11. (Converse of 10) If the angle formed between a line, that is drawn through the end point 
of a chord, and the chord, is equal to the angle subtended by the chord in the alternate segment, then 
the line is a tangent to the circle. 



Proof: 




Consider a circle, with centre O and chord AB. Let line SR pass through point B. Chord AB subtends 
an angle at point Q such that ABS = AQB. 

The aim is to prove that SBR is a tangent to the circle. 

By contradiction. Assume that SBR is not a tangent to the circle and draw XBY such that XBY is a 
tangent to the circle. 



ABX 

However, ABS 

.-.ABX 

But since, ABX 

(9.2) can only be true if, XBS 



AQB (tan-chord theorem) 

AQB (given) 

ABS 

ABS + XBS 





(9.2) 



If XBS is zero, then both XBY and SBR coincide and SBR is a tangent to the circle. 



Exercise 9-7 



1 . Show that Theorem 4 also applies to the following two cases: 



1:52 



CHAPTER 9. GEOMETRY 



9.2 




Q\n More practice (►) video solutions ("fj or help at www.everythingmaths.co. z 



(1.) 022i 



Example 1: Circle Geometry I 



QUESTION 




BD is a tangent to the circle with centre O. 
BO ± AD. 



i:« 



9.2 CHAPTER 9. GEOMETRY 

Prove that: 

1. CFOE is a cyclic quadrilateral 

2. FB = BC 

3. ACOE///ACBF 

4. CD 2 = ED x AD 

r OE_ _ C_D 

BC CO 



SOLUTION 



Step 1 : To show a quadrilateral is cyclic, we need a pair of opposite angles to be 
supplementary, so let's look for that. 



FOE = 90° {BO ± OD) 

FCE = 90° (Z subtended by diameter AE) 

CFOE is a cyclic quadrilateral (opposite Z's supplementary) 

Step 2 : Since these two sides are part of a triangle, we are proving that triangle to be 
isosceles. The easiest way is to show the angles opposite to those sides to be equal. 

Let OEC = x. 

FCB = x (Z between tangent BD and chord CE) 
BFC = x (exterior Z to cyclic quadrilateral CFOE) 
and BF = BC (sides opposite equal Z's in isosceles ABFC) 

Step 3 : To show these two triangles similar, we will need 3 equal angles. We already 
have 3 of the 6 needed angles from the previous question. We need only find the 
missing 3 angles. 

CBF = 180° - 2x (sum of Z's in ABFC) 

OC = OE (radii of circle O) 

.-. ECO = x (isosceles ACOE) 

.-. COE = 180° - 2x (sum of Z's in ACOE) 

• COE = CBF 
. ECO = FCB 

. OEC = CFB 

:. ACOE///ACBF (3 Z's equal) 



Step 4 : This relation reminds us of a proportionality relation between similar triangles. 
So investigate which triangles contain these sides and prove them similar. In this case 3 
equal angles works well. Start with one triangle. 



i;;i 



CHAPTER 9. GEOMETRY 9.2 



In AEDC 



CED = 180° - x (Z's on a straight line AD) 
ECD = 90° — x (complementary Z's) 



Step 5 : Now look at the angles in the other triangle. 

In AADC 

ACT) = 180° - x (sum of Z's ACE and ECD) 
CAD = 90° - x (sum of Z's in ACAE) 



Step 6 : The third equal angle is an angle both triangles have in common. 

Lastly, ADC = EDC since they are the same Z. 

Step 7 : Now we know that the triangles are similar and can use the proportionality 
relation accordingly. 



:. AADC 1 1 1 ACDE (3 Z's equal) 

ED _ CD 
' ' ~CD ~ AD 
:. CD 2 = ED x AD 

Step 8 : This looks like another proportionality relation with a little twist, since not all 
sides are contained in 2 triangles. There is a quick observation we can make about the 
odd side out, OE. 

OE = OC (AOEC is isosceles) 

Step 9 : With this observation we can limit ourselves to proving triangles BOC and 
ODC similar. Start in one of the triangles. 

In ABCO 

OCB = 90° (radius OC on tangent BD) 
CBO = 180° - 2x (sum of Z's in ABFC) 
BOC = 2x- 90° (sum of Z's in ABCO) 



Step 1 : Then we move on to the other one. 

In AOCD 

OCD = 90° (radius OC on tangent BD) 
COD = 180° - 2x (sum of Z's in AOCE) 
CDO = 2x- 90° (sum of Z's in AOCD) 



Step 1 1 : Then, once we've shown similarity, we use the proportionality relation, as 
well as our first observation, appropriately. 



ABOC 1 1 1 AODC (3 Z's equal) 
CO _ CD 
BC ~ CO 

OF 1 CD 

^— = ^— (OE = CO isosceles AOEC) 

BC CO 

135 



9.2 



CHAPTER 9. GEOMETRY 



Example 2: Circle Geometry II 



QUESTION 




FD is drawn parallel to the tangent CB 
Prove that: 

1. FADE is cyclic 

2. AAFE///ACBD 

3 FCxAG _ DCxFE 



SOLUTION 



Step 7 : In this case, the best way to show FADE is a cyclic quadrilateral is to look for 
equal angles, subtended by the same chord. 

Let /LBCD = x 

:. AC AH = x (Z between tangent BC and chord CE) 

.-. Z_FDC = x (alternate Z, FD || CB) 

.-. FADE is a cyclic quadrilateral (chord FE subtends equal Z's) 



Step 2 : To show these 2 triangles similar we will need 3 equal angles. We can use the 
result from the previous question. 



i;s(i 



CHAPTER 9. GEOMETRY 9.2 

Let ZFEA = y 

:. ZFDA = y (Z's subtended by same chord AF in cyclic quadrilateral FADE) 
.-. ZCBD = y (corresponding Z's, FD \\ CB) 
.-. ZFEA = ZCBD 

Step 3 : We have already proved 1 pair of angles equal in the previous question. 

ZBCD = ZFAE (above) 



Step 4 : Proving the last set of angles equal is simply a matter of adding up the angles 
in the triangles. Then we have proved similarity. 



ZAFE = 180° -x-y (Z's in AAFE) 
ZCDB = 180° -x-y (Z's in ACBD) 

:. AAFE///ACBD (3 Z's equal) 

Step 5 : This equation looks like it has to do with proportionality relation of similar 
triangles. We already showed triangles AFE and CBD similar in the previous question. 
So lets start there. 

DC FA 

KD FE 

DC x FE 



BD 



FA 



Step 6 : Now we need to look for a hint about side FA. Looking at triangle CAH we 
see that there is a line FG intersecting it parallel to base CH. This gives us another 
proportionality relation. 

AC FA 

-p— = — =; (FG || CH splits up lines AH and AC proportionally) 

GH r C 

^ A FCxAG 
■• FA =^H— 

Step 7 : We have 2 expressions for the side FA. 

FC.AG _ DC x FE 
' ' GH ~ BD 



137 



9.3 



CHAPTER 9. GEOMETRY 




9.3 Co-ordinate Geometry 




Equation of a Circle 



EMCBZ 



We know that every point on the circumference of a circle is the same distance away from the centre of 
the circle. Consider a point (xi;yi) on the circumference of a circle of radius r with centre at (xo;yo). 



\P(xi;yi) 




Figure 9.3: Circle with centre (x ;yo) and a point P at (x\;yi) 



In Figure 9.3, AOPQ is a right-angled triangle. Therefore, from the Theorem of Pythagoras, we know 
that: 



But, 



OP 2 = PQ 2 + OQ 2 



PQ = Vi- yo 

OQ = xi — xo 

OP = r 

r 2 = (yi - yo) 2 + (si - x ) 2 



But, this same relation holds for any point P on the circumference. Therefore, we can write: 

(x-xo) 2 + (y -yo) 2 = r 1 
for a circle with centre at (xo\ yo) and radius r. 
For example, the equation of a circle with centre (0;0) and radius 4 is: 

(y - yo) 2 + (x- x ) 2 = r 2 



(y-0) 2 + (x-0) 2 = 4 2 
16 



2 . 2 

x +y 



(9.3) 



© See video: VMhrm at www.everythingmaths.co.za 



l:S.s 



CHAPTER 9. GEOMETRY 



Example 3: Equation of a Circle I 



QUESTION 



Find the equation of a circle (centre O) with a diameter between two points, P at (—5; 5) and 
Qat(5;-5). 



SOLUTION 



Step 1 : Draw a picture 

Draw a picture of the situation to help you figure out what needs to be done. 




Step 2 



Find the centre of the circle 

We know that the centre of a circle lies on the midpoint of a diameter. Therefore 
the co-ordinates of the centre of the circle is found by finding the midpoint of the 
line between P and Q. Let the co-ordinates of the centre of the circle be (x ;yo), 
let the co-ordinates of P be (xi;yi) and let the co-ordinates of Q be (x 2 ;y2)- 
Then, the co-ordinates of the midpoint are: 



x 



Vo 



xi + x 2 

2 
-5 + 5 

2 


m + y2 

2 

5 + (-5) 
2 
= 

The centre point of line PQ and therefore the centre of the circle is at (0; 0). 

Step 3 : Find the radius of the circle 

If P and Q are two points on a diameter, then the radius is half the distance 
between them. 



139 



CHAPTER 9. GEOMETRY 



The distance between the two points is: 



- 2 PQ = 


7jV( X 2 -Xl) 2 + (2/2 - J/l) 2 


= 


2 V(5-(-5))H(-5-5^ 


= 


2V(10) 2 + (-10) 2 
1 


= 


-Vioo + ioo 

J200 
V 4 
V50 



Step 4 : Write the equation of the circle 



x + y =50 



Example 4: Equation of a Circle II 



QUESTION 



Find the centre and radius of the circle 
x 2 - Ux + y 2 +Ay = -28. 



SOLUTION 



Step 1 : Change to standard form 

We need to rewrite the equation in the form (x — xq) + {y — j/o) = r 
To do this we need to complete the square 
i.e. add and subtract (| cooefficient of x) 2 and (| cooefficient of y) 2 



Step 2 : Adding cooefficients 

x 2 - lAx + y 2 + 4y = -28 
.-. x 2 - Ux + (7) 2 - (7) 2 + y 2 + Ay + (2) 2 - (2) 2 = -28 

Step 3 : Complete the squares 

:. (i-7) 2 -(7) 2 + (2/ + 2) 2 -(2) 2 = -28 

Step 4 : Take the constants to the other side 

.-. (i-7) 2 -49 + (j/ + 2) 2 -4 = -28 
.-. {x - 7) 2 + (y + 2) 2 = -28 + 49 + 4 
.-. (x-7) 2 + (y + 2) 2 =25 

Step 5 : Read the values from the equation 



110 



CHAPTER 9. GEOMETRY 9.3 



centre is (7; —2) and the radius is 5 units 



Equation of a Tangent to a Circle at a Point ^emcca 
on the Circle 



We are given that a tangent to a circle is drawn through a point P with co-ordinates (xi;yi). In this 
section, we find out how to determine the equation of that tangent. 



(a>i ;ift) 




Figure 9.4: Circle h with centre (x ;yo) has a tangent, g passing through point P at (xi;yi). Line / 
passes through the centre and point P. 

We start by making a list of what we know: 

1. We know that the equation of the circle with centre (xo;yo) and radius r is (2 — x ) 2 + (y— yo) 2 = 
r 2 . 

2. We know that a tangent is perpendicular to the radius, drawn at the point of contact with the 
circle. 

As we have seen in earlier grades, there are two steps to determining the equation of a straight line: 

Step 1 : Calculate the gradient of the line, m. 
Step 2: Calculate the ^-intercept of the line, c. 

The same method is used to determine the equation of the tangent. First we need to find the gradient 
of the tangent. We do this by finding the gradient of the line that passes through the centre of the circle 
and point P (line / in Figure 9.4), because this line is a radius and the tangent is perpendicular to it. 

2/i - Vo ,„ ., 

mr = (9.4) 

Xi - xo 

The tangent (line g) is perpendicular to this line. Therefore, 

rrif X m 9 = —1 

So, 

1 

m„ = 

mj 

141 



9.3 



CHAPTER 9. GEOMETRY 



Now, we know that the tangent passes through (xi; yi) so the equation is given by: 

; - xi) 
-(x - xi) 



y - 2/1 = m g (x — xi) 

1 

y -2/i = 



771/ 

y i/l yi-yo *- 1 ' 

xi-xo 
Xi - x , 



y-yi 



yi - 2/0 



1 / N 

-(x - Xl) 



For example, find the equation of the tangent to the circle at point (1: 1). The centre of the circle is at 
(0; 0). The equation of the circle is x 2 + y 2 = 2. 



Use 



2/ -2/i 



xi - x a 
2/i - 2/o 



(x - Xi) 



with(xo;j/o) = (0:0)and(xi;j/i) = (1; 1). 



2/ -2/1 


— 


: -(x- 

2/1 -2/o 


2/-1 


= 


(x — 1) 

1-0 V ' 


2/-1 


= 


-j(*"l) 


y 


= 


-(x-l) + l 


y 


= 


-x + 1 + 1 


V 


= 


-x + 2 



Exercise 9-8 



1. Find 

(a) 
(b) 
(0 
(d) 
(e) 

2. (a) 

(b) 
(c) 

3. (a) 
(b) 
(0 

4. Find 

(a) 
(b) 
(0 
(d) 
(e) 



the equation of the circle: 

with centre (0; 5) and radius 5 
with centre (2; 0) and radius 4 
with centre (5; 7) and radius 18 
with centre (—2: 0) and radius 6 
with centre (—5; —3) and radius \/3 

Find the equation of the circle with centre (2; 1) which passes through (4; 1). 
Where does it cut the line y = x + 1? 
Draw a sketch to illustrate your answers. 

Find the equation of the circle with centre (—3: —2) which passes through (1; —4). 
Find the equation of the circle with centre (3; 1) which passes through (2; 5). 
Find the point where these two circles cut each other. 

the centre and radius of the following circles: 

(x - 9) 2 + (y- 6) 2 = 36 
(x - 2) 2 + (y - 9) 2 = 1 
(x + 5) 2 + (y + 7) 2 = 12 
(x + 4) 2 + (y + 4) 2 = 23 
3(x-2) 2 + 3(i/ + 3) 2 = 12 



1 12 



CHAPTER 9. GEOMETRY 



9.4 



(f) x 2 - 3x + 9 = y 2 + 5y + 25 = 17 
5. Find the re and y intercepts of the following graphs and draw a sketch to illustrate your answer: 



(a) (x + 7) 2 + (y-2) 2 = 8 

(b) x 2 + (y - 6) 2 = 100 

(c) {x + if + y 2 = 16 

(d) {x - 5) 2 + (y + l) 2 = 25 

6. Find the centre and radius of the following circles: 

(a) x 2 + 6x + y 2 - 12y = -20 

(b) x 2 + 4x + y 2 - %y = 

(c) x 2 + y 2 + % = 7 

(d) x 2 - Qx + y 2 = 16 

(e) x 2 - 5x + y 2 + 3y = -| 

(f) x 2 - %nx + y 2 + lOny = 9n 2 

7. Find the equation of the tangent to each circle: 

(a) x 2 +y 2 = 17 at the point (1; 4) 

(b) x 2 +y 2 = 25 at the point (3; 4) 

(c) (x + l) 2 + (y - 2) 2 = 25 at the point (3; 5) 

(d) (x - 2) 2 + (y- l) 2 = 13 at the point (5; 3) 



f/Vy More practice f ►) video solutions f Oj or help at www.everythingmaths.co.za 



(1.)01gt (2.)01gu (3.)01gv (4.)01gw (5.)01gx (6.) 01gy 
(7.)01gz 




9.4 Transformations 




Rotation of a Point About an Angle 9 



EMCCC 



First we will find a formula for the co-ordinates of P after a rotation of ( 



We need to know something about polar co-ordinates and compound angles before we start. 



113 



CHAPTER 9. GEOMETRY 



Polar co-ordinates 



Notice that sin a = - : .y = r sin a 
and cos a = — .'. x = r cos a 
so P can be expressed in two ways: 

P(x:y) rectangular co-ordinates 
or P(r; a) polar co-ordinates. 



Compound angles 

(See derivation of formulae in Chapter 12) 

sin (a + f3) 
cos (a + f3) 



sin a cos f) + sin /3 cos a 
cos a cos P — sin a sin /? 



Now consider P' after a rotation of ( 



P(x; y) = P(r cos ct; r sin a) 
P (r cos (a + #);r sin (a + 8)) 

Expand the co-ordinates of P' 



co-ordinate 



y — co-ordinate 



rcos(a + 8) 

r [cos a cos 8 — sin a sin 6] 
r cos a cos 6 — r sin a sin # 
x cos 8 — y sin 6 

r sin (a + 6) 

r [sin a cos # + sin 8 cos a] 
r sin a cos 8 + r cos a sin 
i/ cos 8 + x sin # 



which gives the formula P' = [(x cos 8 — y sin 8; y cos 6 + x sin #)] 





p' 


e 


a \ 


P = 


(rcosa; 















So to find the co-ordinates of P(l; %/3) after a rotation of 45°, we arrive at: 

P = [(xcosd — ysm8);(ycos8 + xs'mS)] 

= [(lcos45° - V3sin45°);(V3cos45° + 1 sin 45°) 

,-\/2 v/2/'\V2 V2 
1- V3 y^+1 

^2 ' ^2 



Exercise 9-9 



114 



CHAPTER 9. GEOMETRY 



9.4 



Any line OP is drawn (not necessarily in the first 
quadrant), making an angle of 9 degrees with the 
x-axis. Using the co-ordinates of P and the angle 
a, calculate the co-ordinates of P', if the line OP 
is rotated about the origin through a degrees. 





P 


a 


1. 


(2; 6) 


60° 


2. 


(4; 2) 


30° 


3. 


(5;-i) 


45° 


4. 


(-3; 2) 


120° 


5. 


(-4;-i) 


225° 


6. 


(2; 5) 


-150° 



o 



More practice ( ►) video solutions (cj or help at www.everythingmaths.co.za 



(1.)022j 



Characteristics of Transformations 



EMCCD 



Rigid transformations like translations, reflections, rotations and glide reflections preserve shape and 
size, and that enlargement preserves shape but not size. 



Activity: 



Geometric Transformations 



Draw a large 15x15 grid and plot 
AABC with A(2;6), B(5;6) and 
C(5; 1). Fill in the lines y = x and 

y = -x. 

Complete the table below , by draw- 
ing the images of AABC under the 
given transformations. The first one 
has been done for you. 







15 












10 










5 


A(2;6) B(5;6)' 




















7(5; 1) 




-15 


-10 
' y = x 


—5 
-10 


5 10 

'•., *„, 

'.'I 

/ 1 

*- - -I 

A' B' 

y = -I ■ 


15 






-15 









115 



9.4 



CHAPTER 9. GEOMETRY 



Transformation 


Description 
(translation, reflection, 
rotation, enlargement) 


Co-ordinates 


Lengths 


Angles 


(x;y) -> (x;-y) 


reflection about the x-axis 


A' (2; -6) 
B'(5;-6) 

C"(5;-2) 


A'B' = 3 
B'C*' = 4 
A'C = 5 


B' = 90° 

tan A = 4/3 

.-. i = 53°,C , = 37° 


(x;y) -»■ (x + 1; 1/ - 2) 










(a?;y) -> (-a;;i/) 










(x;y) -> (-i/;x) 










(ar;y) -► (-a; -y) 










(x;y) -+ (2x;2y) 










(x;y) -* (y;x) 










(x;y) -t(y;x + 1) 











A transformation that leaves lengths and angles unchanged is called a rigid transformation. 
Which of the above transformations are rigid? 



Chapter 9 


End of Chapter Exercises 



1 . AABC undergoes several transformations forming AA'B'C Describe the relation- 
ship between the angles and sides of AABC and AA'B'C' (e.g., they are twice as 
large, the same, etc.) 



Transformation 


Sides 


Angles 


Area 


Reflect 








Reduce by a scale factor of 3 








Rotate by 90° 








Translate 4 units right 








Enlarge by a scale factor of 2 









2. ADEFhasE = 30°, DE = 4 cm, EF = 5 cm. ADEF is enlarged by a scale factor 
of 6 to form AD'E'F'. 

(a) Solve ADEF 

(b) Hence, solve AD'E'F' 

3. AXYZ has an area of 6cm 2 . Find the area of AX'Y'Z' if the points have been 
transformed as follows: 

(a) (x,y)^(x + 2;y + 3) 

(b) (x,y) -> (y;x) 



ll(i 



CHAPTER 9. GEOMETRY 9.4 



(c) (a; 


,y) 


-► (4x: 


J/) 


(d) (x 


,y) 


-+ (3a;: 


2/ + 2) 


(e) (x 


,y) 


— > (—a 


•;-?/) 


0) (x 


,v) 


-► (a;;- 


-2/ + 3) 


(g) (a: 


,y) 


-> (4s; 


%) 


(h) (x 


,y) 


^(-3x;4y) 



ft" 1 ) More practice (►) video solutions f9) or help at www.everythingmaths 
(1.) OlhO (2.) Olhl (3.)01h2 



147 



Trigonometry 





7 0. 1 Introduction 




In this chapter we will build on the trigonometry from previous years by looking at the result applying 
trigonometric functions to sums and differences of angles. We will also apply trigonometry to solving 
geometric problems in two and three dimensions. 

© See introductory video: VMhtu at www.everythingmaths.co.za 




10.2 Compound Angle Identities 




Derivation of sin (a + (5) 



EMCCG 



We have, for any angles a and 0, that 



sin(a + 0) = sin a cos + sin /3 cos a 



How do we derive this identity? It is tricky, so follow closely. 



Suppose we have the unit circle shown below. The two points L(a; b) and K(x; y) are on the circle. 



148 



CHAPTER 10. TRIGONOMETRY 



10.2 



K(x;y) 



L(a:b) 




We can get the coordinates of L and K in terms of the angles a and 0. For the triangle LOM, we have 
that 



sin /3 = — 

a a 

cos p = — 



b = sin /3 
a = cos /3 



Thus the coordinates of L are (cos /3; sin /3). In the same way as above, we can see that the coordinates 
of K are (cos a; sin a). The identity for cos(a — /3) is now determined by calculating KI? in two ways. 
Using the distance formula (i.e. d = \J\x~i — ii) 2 + (2/2 — 3/1) 2 or d 2 = (x 2 — xi) 2 + (y 2 — yi) 2 ), we 
can find KL 2 : 

KL = (cos a — cos/3) + (sin a — sin/3) 

= cos a — 2 cos a cos /3 + cos /3 + sin a — 2 sin a sin /3 + sin /3 
= (cos a + sin a) + (cos /3 + sin /3) — 2 cos a cos /3 — 2 sin a sin /3 
= 1 + 1 — 2(cos a cos /3 + sin a sin /3) 
= 2 — 2(cosacos/3 + sinasin/3) 
The second way we can determine KL 2 is by using the cosine rule for AKOL: 
KL 2 = KO 2 + LO 2 -2.KO.LO. cos(a - 0) 
= 1 2 + l 2 -2(l)(l)cos(a-/3) 
= 2 - 2 . cos(a - 13) 
Equating our two values for KL 2 , we have 

2 — 2 . cos(a — /3) = 2 — 2(cosacos/3 + sin a . sin/3) 
^=> cos(a — /3) = cos a . cos /3 + sin a . sin /3 
Now let a — > 90° — a. Then 

cos(90° - a - /3) = cos(90° - a) cos /3 + sin(90° -a)sin/3 
= sin a . cos /3 + cos a . sin /3 
But cos(90° - (a + /3)) = sin(a + /3). Thus 

sin(a + /3) = sin a . cos /3 + cos a . sin /3 



J. J.<) 



70.2 



CHAPTER 10. TRIGONOMETRY 



Derivation of sin (a - (3) 



EMCCH 



We can use 
to show that 
We know that 
and 
Therefore, 



sin(a + 0) = sin a cos + cos a sin 

sin(a — 0) = sin a cos — cos a sin 

sin(-fl) = -sin(0) 

cos(— 0) = COS0 

sin(a - /3) = sin(a + (-£)) 

= sinacos(— 0) + cosasin(— /3) 
= sin a cos /3 — cos a sin /3 



Derivation of cos (a + /5) 



EMCCI 



We can use 
to show that 
We know that 
Therefore, 



sin(a — 0) = sin a cos — sin cos a 

cos(a + 0) = cos a cos — sin a sin 

sin(fl) = cos(90-6»). 

cos(a + /3) = sin(90- (a + /3)) 

= sin((90 - a) - 0)) 

= sin(90 — a) cos/3 — sin/3cos(90 — a) 

= cos a cos — sin /3 sin a 



Derivation of cos (a - /3) 



EMCCJ 



We found this identity in our derivation of the sin(o + 0) identity. We can also use the fact that 

sin(ct + 0) = sin a cos + cos a sin 
to derive that 



As 



cos(ct — 0) = cos a cos + sin a sin 
cos(0) = sin(90-6»), 

150 



CHAPTER 10. TRIGONOMETRY 



10.2 



we have that 



cos(ct -P) = sin(90 - (a - /?)) 

= sin((90 - a) + /?)) 

= sin(90 — q) cos + cos(90 — a) sin /3 

= cos a cos f) + sin a sin /3 



Derivation of sin 2a 



EMCCK 



We know that 



When a = /9, we have that 



sin(a + /3) = sin « cos /3 + cos a sin /3 



sin(2a) = sin(a + a) = sin a cos a + cos a sin a 
= 2 sin a cos a 



Derivation of cos 2a 



EMCCL 



We know that 



When a = /}, we have that 



cos(a + ft) = cos a cos /3 — sin a sin /3 



cos(2a) = cos(q + a) = cos a cos a — sin a sin a 



cos a — sin a 



However, we can also write 
and 

by using 



Activity: 



cos 2o = 2 cos a — 1 



cos 2a = 1 — 2 sin a 



sin a + cos a = 1 . 



77ie cos 2a Identity 



Use 



to show that: 



151 



70.2 



CHAPTER 10. TRIGONOMETRY 



Problem-solving Strategy for Identities 



EMCCM 



Tip 


When 


proving trigono- 


metric 


identities, never 


assume that the [eft hand 


side is 


equal to the right 


hand side. You need to 


show that both sides are 


equal. 





The most important thing to remember when asked to prove identities is: 

A suggestion for proving identities: It is usually much easier simplifying the more complex side of an 
identity to get the simpler side than the other way round. 



Example 1: Trigonometric Identities 1 



QUESTION 


Prove that sin 75° = (v ^ 3+1) without using a calculator. 


SOLUTION 




Step 1 : Identify a strategy 

We only know the exact values of the trig functions for a few special angles (30°; 


45°; 60°; etc.). We 


can see that 75° = 30° + 45°. Thus we can use our double- 


angle identity for sin(a + 0) to express sin 75° in terms of known trig function 
values. 


Step 2 : Execute strategy 




sin 75° = 


sin(45°+30°) 


= 


sin(45°) cos(30°) + cos(45°) sin(30°) 


= 


1 ^3 1 1 
v/2' 2 + ^2'2 




73 + 1 




2^2 




n/3 + 1 V2 




2s/2 '" s/2 




v/2(\/3 + l) 

4 





lo2 



CHAPTER 10. TRIGONOMETRY 10.2 



Example 2: Trigonometric Identities 2 



QUESTION 



Deduce a formula tor tan(a + 0) in terms of tana and tan/3. 
Hint: Use the formulae tor sin(a + 0) and cos(a + 0) 



SOLUTION 



Step 1 : Identify a strategy 

We can express tan(a + 0) in terms of cosines and sines, and then use the 
double-angle formulae for these. We then manipulate the resulting expression in 
order to get it in terms of tana and tan/3. 



Step 2 : Execute strategy 

tan(a + /3) 



sin(a + /3) 

cos(a + /3) 

sin a . cos + cos a . sin /3 

cos a . cos — sin a . sin 

sin a . cos j3 ■ cos a . sin /3 
cos a . cos cos a . cos /3 



tan a + tan 
1 — tana . tan/3 



Example 3: Trigonometric Identities 3 



QUESTION 



Prove that 

sin 6 + sin 20 

— - = tan 

1 + cos 9 + cos 20 

In fact, this identity is not valid for all values of 9. Which values are those? 



SOLUTION 



Step 1 : Identify a strategy 

The right-hand side (RHS) of the identity cannot be simplified. Thus we should 
try simplify the left-hand side (LHS). We can also notice that the trig function on 
the RHS does not have a 29 dependence. Thus we will need to use the double- 
angle formulae to simplify the sin 20 and cos 29 on the LHS. We know that tan# 



l.VS 



70.2 CHAPTER 10. TRIGONOMETRY 



is undefined for some angles 9. Thus the identity is also undefined for these 9, 
and hence is not valid for these angles. Also, for some 9, we might have division 
by zero in the LHS, which is not allowed. Thus the identity won't hold for these 
angles also. 



Step 2 : Execute the strategy 

sin 9 + 2 sin 9 cos I 
LHS 



1 + cos 9 + (2 cos 2 9 - 1) 
sin6»(l + 2cos#) 
cos6»(l + 2cos6>) 
sin 9 
cos 9 
tan 9 

RHS 



We know that tan 6 is undefined when 9 = 90° + 180°n, where n is an integer. 
The LHS is undefined when 1 + cos 9 + cos 29 = 0. Thus we need to solve this 
equation. 

1 + cos 9 + cos 29 = 
=>• cos6»(l + 2cos6») = 



The above has solutions when cos9 = 0, which occurs when 9 = 90° + 180°n, 
where n is an integer. These are the same values when tan# is undefined. It 
also has solutions when 1 + 2 cos 9 = 0. This is true when cos 9 = — \, and thus 
9 = ... - 240°, -120°, 120°, 240°, .... To summarise, the identity is not valid 
when 9 = ... - 270° ; -240° ; - 120° ; -90° ; 90° ; 120° ; 240° ; 270° ; . . . 



Example 4: Trigonometric Equations 



QUESTION 



Solve the following equation for y without using a calculator: 

1 — sin y — cos 2y 
sin 2y — cos y 



SOLUTION 



Step 1 : Identify a strategy 

Before we are able to solve the equation, we first need to simplify the left-hand 
side. We do this by using the double-angle formulae. 



Step 2 : Execute the strategy 



lol 



CHAPTER 10. TRIGONOMETRY 



10.3 



1 — sin y — (1 — 2 sin y) 

2 sin y cos i/ — cos y 

2 sin y — sin y 

cos 1/(2 sin y — 1) 

sinj/(2sini/ — 1) 

cosy(2sinj/ — 1) 

tan y 

/ = 135° + 180° n; n e Z 



-1 

-1 

-1 
-1 




10.3 Applications of Trigonometric 
Functions 




Problems in Two Dimensions 



EMCCO 



Example 5: Problem in Two Dimensions 



QUESTION 








A 






o 1 




I ° 
















X / N^~~~--\^^ 








c *^- — ■ 


B 




Tor the figure below, we are given that 


CD = BD = x. 








Show that BC 2 = 2x 2 ( 1 + sin 6) . 









ir>r> 



70.3 



CHAPTER 10. TRIGONOMETRY 




SOLUTION 



Step 7 : Identify a strategy 

We want CB, and we have CD and BD. If we could get the angle BDC, 
then we could use the cosine rule to determine BC. This is possible, as AABD 
is a right-angled triangle. We know this from circle geometry, that any triangle 
circumscribed by a circle with one side going through the origin, is right-angled. 
As we have two angles of AABD, we know ADB and hence BDC. Using the 
cosine rule, we can get BC 2 . 



Step 2 : Execute the strategy 



Thus 



ADB = 180° 



BDC 



90 



: 90° - e 



180° - ADB 
= 180° - (90° - 0) 
= 90° + 

Now the cosine rule gives 

BC 2 = CD 2 + BD 2 -2.CD.BD. cos(BDC) 

= x 2 + x 2 - 2 . x 2 . cos(90° + 0) 

= 2x 2 + 2x 2 [sin(90°) cos(0) + sin(0) cos(90°)] 

= 2x 2 + 2x 2 [ 1 . cos(0) + sin(6>) . 0] 

= 2x 2 (l-sin6») 



Exercise 10-1 



1 . For the diagram on the right, 



l"i(i 



CHAPTER 10. TRIGONOMETRY 



10.3 



(a) Find AOC in terms of 9 

(b) Find an expression for: 



cosy 
sin 9 
sin 26» 




(c) Using the above, show that sin 2$ 
2 sin 9 cos 9. 

(d) Now do the same for cos 29 and tan 6. 



2. DC is a diameter of circle O with radius r. CA = r, AB = DE and DOE ■ 
Show that cos 9 = |, 

E 




3. The figure below shows a cyclic quadrilateral with ^ = ^|-. 

(a) Show that the area of the cyclic quadrilateral \s DC .DA. sinD. 

(b) Find expressions for cosD and cosB in terms of the quadrilateral sides. 

(c) Show that 2CA 2 = CD 2 + DA 2 + AB 2 + BC 2 . 

(d) Suppose that BC = 10, CD = 15, AD = 4 and AB = 6. Find CA 2 . 

(e) Find the angle D using your expression for cos D. Hence find the area of ABCD. 




Q^j More practice f ►) video solutions (7) or help at www.everythingmaths.co.za 



(l.)01ai (2.)01aj (3.) Olak 



157 



70.3 



CHAPTER 10. TRIGONOMETRY 



Problems in 3 Dimensions 



EMCCP 



Example 6: Height of tower 



QUESTION 




D is the top of a tower of height h. Its base is at A. The triangle ABC lies on the ground (a 
horizontal plane). If we have that BC = b, DBA = a, DBC = and DCB = 9, show that 



h- 



b sin q sin 9 

sin(/3 + 9) 




SOLUTION 



1").S 



CHAPTER 10. TRIGONOMETRY 10.3 



Step 1 : Identify a strategy 

We have that the triangle ABD is right-angled. Thus we can relate the height 
h with the angle a and either the length BA or BD (using sines or cosines). 
But we have two angles and a length for ABCD, and thus can work out all the 
remaining lengths and angles of this triangle. We can thus work out BD. 

Step 2 : Execute the strategy 

We have that 

h 
-=—- = sin a 
BD 

=*> h = BDsvnct 

Now we need BD in terms of the given angles and length b. Considering the 
triangle BCD, we see that we can use the sine rule. 

sin sm(BDC) 

~BD 6 



BD 



But BDC = 180° -13-9, and 



bsmd 
sm{BDC) 



sin(180° -/3-0) = - sin(-/3 - 
= sin(/3 + &) 



So 

BD 



b sin 6 
sm(BDC) 

bsmd 
sin(/3 + 9) 
BD sin a 
b sin a sin 9 
sin(/3 + 9) 



Exercise 10-2 



1 . The line BC represents a tall tower, with B at its foot. Its angle of elevation from D is 9. We are 
also given that BA = AD = x. 



159 



70.4 



CHAPTER 10. TRIGONOMETRY 



C 




D 



2. Find the height of the tower BC in terms of x, tan 61 and cos 2a. 

3. Find BC if we are given that x = 140 m, a = 21° and 9 = 9°. 



I\*j More practice f ►) video solutions CfJ or help at www.everythingmaths.co.za 



(1.)01am (2.)01an (3.) 01ap 




10.4 Other Geometries 




Taxi cab Geometry 



EMCCR 



Taxicab geometry, considered by Hermann Minkowski in the 19th century, is a form of geometry in 
which the usual metric of Euclidean geometry is replaced by a new metric in which the distance 
between two points is the sum of the (absolute) differences of their coordinates. 



Manhattan Distance 



EMCCS 



The metric in taxi-cab geometry, is known as the Manhattan distance, between two points in an 
Euclidean space with fixed Cartesian coordinate system as the sum of the lengths of the projections of 
the line segment between the points onto the coordinate axes. 



Kid 



CHAPTER 10. TRIGONOMETRY 



10.4 



For example, the Manhattan distance between the point Pi with coordinates (xi;yi) and the point P 2 
at {xr,yi) is 



|xi -x 2 \ + \yi -yi\ 



(10.1) 



' • • • • ■ • i m • • • *~*~i 
! I 

• > 

, , 



Figure 10.1: Manhattan distance (dotted and solid) compared to Euclidean distance (dashed). In each 
case the Manhattan distance is 12 units, while the Euclidean distance is a/36 



The Manhattan distance changes if the coordinate system is rotated, but does not depend on the 
translation of the coordinate system or its reflection with respect to a coordinate axis. 



Manhattan distance is also known as city block distance or taxi-cab distance. It is given these names 
because it is the shortest distance a car would drive in a city laid out in square blocks. 



Taxicab geometry satisfies all of Euclid's axioms except for the side-angle-side axiom, as one can 
generate two triangles with two sides and the angle between them the same and have them not be 
congruent. In particular, the parallel postulate holds. 



A circle in taxicab geometry consists of those points that are a fixed Manhattan distance from the 
centre. These circles are squares whose sides make a 45° angle with the coordinate axes. 



161 



70.4 



CHAPTER 10. TRIGONOMETRY 




Summary of the Trigonometric 
Rules and Identities 




Pythagorean Identity 
cos 2 9 + sin 2 9 = 1 



Cofunction Identities 

sin(90° - 8) = cos 9 
cos(90° -8) = sin8 



Ratio Identities 



tan 8 ■ 



sin 9 
cos 



Odd/Even Identities 

sin(— 8) = — sin 8 
cos(—8) = cos 8 
tan(— 8) = —tan 8 



Periodicity Identities Double Angle Identities 



sin(6»±360°) = sinfl 
cos(0±36O°) = cos 6» 
tan(6>±180°) = tan6» 



sm(28) = 2 sin0 cos 8 
cos (28) = cos 2 8 — sin 2 
cos (261) = 1 -2sin 2 
tan (26))= 1 a '" l ° 



Addition/Subtraction Identities 



Area Rule 



Cosine rule 



sin(6» + ^ 
sin (8 — <j> 
cos (9 + 4> 
cos (8 — <f> 
tan (8 + <p 
tan (8 — (j) 



■ sin U cos <p + cos o sin q> 

■ sin 8 cos </> — cos 8 sin 
= cos 8 cos </> — sin 8 sin 
= cos 8 cos + sin 8 sin </> 

tan <p+tan 
1— tan 6 tan ci 
tan <A — tan 



Area = \bc sin A 
Area = | ah sin C 
Area = \ac sin B 



a 2 = 6 2 + c 2 - 2bccosA 
6 2 = a 2 + c 2 - 2ac cos B 
c 2 = a 2 + b 2 -2abcosC 



1+tan 8 tan <f> 



Sine Rule 



Chapter 10 



End of Chapter Exercises 



Do the following without using a calculator. 

1. Suppose cosd = 0,7. Find cos 29 and cos id. 

2. If sin 8 = |, again find cos 29 and cos 46 1 . 

3. Work out the following: 

(a) cos 15° 

(b) cos 75° 

(c) tan 105° 

(d) cos 15° 

(e) cos 3° cos 42° — sin 3° sin 42° 

(f) l-2sin 2 (22,5°) 

4. Solve the following equations: 

(a) cos 39 . cos 9 — sin 39 . sin 9 = — \ 



11,2 



CHAPTER 10. TRIGONOMETRY 10.4 



(b) 3 sin 8 = 2cos 2 6> 

5. Prove the following identities 

(a) Sin 3 6 = 3 si- 9 -sin 38 

(b) cos 2 q(1 — tan 2 a) = cos la 

(c) 4 sin 8 . cos 8 . cos 26* = sin 48 

(d) 4 cos 3 x — 3 cos x = cos 3x 

/ \ i sin lit 

(e) tan !/ = — „ , . 

6. (Challenge question!) If a + b + c = 180°, prove that 

sin a+sin 6+sin c = 3cos(a/2) cos(6/2) cos(c/2)+cos(3a/2) cos(36/2) cos(3c/2) 



A"y More practice f ►) video solutions \jf) or help at www.everythingmaths.co.za 
(1.)01aq (2.)01ar (3.) 01as (4.) 01 at (5.) Olau (6.) 01av 



l(i:i 



Statistics 




1 1 .1 Introduction 



EMCCU 



In this chapter, you will use the mean, median, mode and standard deviation of a set of data to identify 
whether the data is normally distributed or whether it is skewed. You will learn more about populations 
and selecting different kinds of samples in order to avoid bias. You will work with lines of best fit, and 
learn how to find a regression equation and a correlation coefficient. You will analyse these measures 
in order to draw conclusions and make predictions. 

© See introductory video: VMhwi at www.everythingmaths.co.za 




7 7.2 Normal Distribution 




Activity: 



You are given a table of data below. 



75 


67 


70 


71 


71 


73 


74 


75 


80 


75 


77 


78 


78 


78 


78 


79 


91 


81 


82 


82 


83 


86 


86 


87 



1 . Calculate the mean, median, mode and standard deviation of the data. 

2. What percentage of the data is within one standard deviation of the mean? 

3. Draw a histogram of the data using intervals 60 < x < 64; 64 < x < 68; etc. 

4. Join the midpoints of the bars to form a frequency polygon. 



If large numbers of data are collected from a population, the graph will often have a bell shape. If the 
data was, say, examination results, a few learners usually get very high marks, a few very low marks 
and most get a mark in the middle range. We say a distribution is normal if 



the mean, median and mode are equal. 



it is symmetric around the mean. 



±68% of the sample lies within one standard deviation of the mean, 95% within two standard 
deviations and 99% within three standard deviations of the mean. 



164 



CHAPTER 11. STATISTICS 



11.2 




x + a x + 2cr x + 3a 



What happens if the test was very easy or very difficult? Then the distribution may not be symmetrical. 
If extremely high or extremely low scores are added to a distribution, then the mean tends to shift 
towards these scores and the curve becomes skewed. 



If the test was very difficult, the mean score is shifted to the 
left. In this case, we say the distribution is positively skewed, 
or skewed right. 



If it was very easy, then many learners would get high scores, 
and the mean of the distribution would be shifted to the right. 
We say the distribution is negatively skewed, or skewed left. 




Skewed right 




Skewed left 



Exercise 11-1 



1. Given the pairs of normal curves below, sketch the graphs on the same set of axes and show 
any relation between them. An important point to remember is that the area beneath the curve 
corresponds to 100%. 



(a) Mean = 8, standard deviation = 4 and Mean 

(b) Mean = 8, standard deviation = 4 and Mean 

(c) Mean = 8, standard deviation = 4 and Mean 



4, standard deviation = 
16, standard deviation 
8, standard deviation = 



8 



2. After a class test, the following scores were recorded: 



Test Score 


Frequency 


3 


1 


4 


7 


5 


14 


6 


21 


7 


14 


8 


6 


9 


1 


Total 


64 


Mean 


6 


Standard Deviation 


1,2 



(a) Draw the histogram of the results. 

(b) Join the midpoints of each bar and draw a frequency polygon. 



lfi.") 



11.3 



CHAPTER 11. STATISTICS 



(c) What mark must one obtain in order to be in the top 2% of the class? 

(d) Approximately 84% of the pupils passed the test. What was the pass mark? 

(e) Is the distribution normal or skewed? 

3. In a road safety study, the speed of 175 cars was monitored along a specific stretch of highway 
in order to find out whether there existed any link between high speed and the large number of 
accidents along the route. A frequency table of the results is drawn up below. 



Speed (km.h 1 ) 


Number of cars (Frequency) 


50 


19 


60 


28 


70 


23 


80 


56 


90 


20 


100 


16 


110 


8 


120 


5 



The mean speed was determined to be around 82 km-hr 1 whi le the median speed was worked 
out to be around 84,5 km-hr -1 . 

(a) Draw a frequency polygon to visualise the data in the table above. 

(b) Is this distribution symmetrical or skewed left or right? Give a reason fro your answer. 



f/Vj More practice Crj video solutions (9 J or help at www.everythingmaths.co.za 



(1.)01aw (2.) 01 ax (3.) 01 ay 




11.3 Extracting a Sample 
Population 




Suppose you are trying to find out what percentage of South Africa's population owns a car. One way 
of doing this might be to send questionnaires to peoples homes, asking them whether they own a car. 
However, you quickly run into a problem: you cannot hope to send every person in the country a 
questionnaire, it would be far to expensive. Also, not everyone would reply. The best you can do is 
send it to a few people, see what percentage of these own a car, and then use this to estimate what 
percentage of the entire country own cars. This smaller group of people is called the sample popula- 
tion. 

The sample population must be carefully chosen, in order to avoid biased results. How do we do 

this? 

First, it must be representative. If all of our sample population comes from a very rich area, then almost 

all will have cars. But we obviously cannot conclude from this that almost everyone in the country has 

a car! We need to send the questionnaire to rich as well as poor people. 

Secondly, the size of the sample population must be large enough. It is no good having a sample 

population consisting of only two people, for example. Both may very well not have cars. But we 

obviously cannot conclude that no one in the country has a car! The larger the sample population 

size, the more likely it is that the statistics of our sample population corresponds to the statistics of the 

entire population. 



166 



CHAPTER 11. STATISTICS 



11.4 



So how does one ensure that ones sample is representative? There are a variety of methods avail- 
able, which we will look at now. 



Random Sampling. Every person in the country has an equal chance of being selected. It is 
unbiased and also independent, which means that the selection of one person has no effect on 
the selection on another. One way of doing this would be to give each person in the country a 
number, and then ask a computer to give us a list of random numbers. We could then send the 
questionnaire to the people corresponding to the random numbers. 

Systematic Sampling. Again give every person in the country a number, and then, for example, 
select every hundredth person on the list. So person with number 1 would be selected, person 
with number 100 would be selected, person with number 200 would be selected, etc. 

Stratified Sampling. We consider different subgroups of the population, and take random sam- 
ples from these. For example, we can divide the population into male and female, different ages, 
or into different income ranges. 

Cluster Sampling. Here the sample is concentrated in one area. For example, we consider all 
the people living in one urban area. 



Exercise 11-2 



1 . Discuss the advantages, disadvantages and possible bias when using 

(a) systematic sampling 

(b) random sampling 

(c) cluster sampling 

2. Suggest a suitable sampling method that could be used to obtain information on: 

(a) passengers views on availability of a local taxi service. 

(b) views of learners on school meals. 

(c) defects in an item made in a factory. 

(d) medical costs of employees in a large company. 

3. 2% of a certain magazines' subscribers is randomly selected. The random number 16 out of 50, 
is selected. Then subscribers with numbers 16; 66; 116; 166; ... are chosen as a sample. What 
kind of sampling is this? 



More practice CrJ video solutions C'?) or help at www.everythingmaths 



(l.)Olaz (2.)01b0 (3.) Olbl 




7 1 .4 Function Fitting and 
Regression Analysis 



In Grade 1 1 we recorded two sets of data (bivariate data) on a scatter plot and then we drew a line of 
best fit as close to as many of the data items as possible. Regression analysis is a method of finding 



107 



7 7.4 



CHAPTER 11. STATISTICS 



out exactly which function best fits a given set of data. We can find out the equation of the regression 
line by drawing and estimating, or by using an algebraic method called "the least squares method", 
available on most scientific calculators. The linear regression equation is written y = a + bx (we say 
y-hat) or y = A + Bx. Of course these are both variations of a more familiar equation y = mx + c. 

Suppose you are doing an experiment with washing dishes. You count how many dishes you begin 
with, and then find out how long it takes to finish washing them. So you plot the data on a graph of 
time taken versus number of dishes. This is plotted below. 




* d 



2 3 4 5 

Number of dishes 



If t is the time taken, and d the number of dishes, then it looks as though t is proportional to d, ie. 
t = m .d, where m is the constant of proportionality. There are two questions that interest us now. 

1 . How do we find m? One way you have already learnt, is to draw a line of best-fit through the 
data points, and then measure the gradient of the line. But this is not terribly precise. Is there a 
better way of doing it? 

2. How well does our line of best fit really fit our data? If the points on our plot don't all lie close to 
the line of best fit, but are scattered everywhere, then the fit is not "good", and our assumption 
that t = m .d might be incorrect. Can we find a quantitative measure of how well our line really 
fits the data? 

In this chapter, we answer both of these questions, using the techniques of regression analysis. ® See 
simulation: VMibv at www.everythingmaths.co.za) 



Example 1: Fitting by hand 



QUESTION 


Use 
the I 

X 


the da 
ine thi 

1,0 


ta give 
\t best 

2,4 


n to d 
seems 

3,1 


r aw a . 
to fit 

4,9 


catter 
hedat 

5,6 


plot and line of best fit. Now write down the equation of 
a. 

6,2 


y 


2,5 


2,8 


3,0 


4,8 


5,1 


5,3 



k;n 



CHAPTER 11. STATISTICS 



11.4 



SOLUTION 


Step 1 : Drawing the graph 


The first step is to draw the graph. This is shown below. 

V 


7 - 




6 - 


^ 


5 - 


• ^-*^"^ 


4 - 
3 - 


^-^ 






• ^^""^ 1 


2 - 


^^ \ 


1 - 




n 


1 1 1 1 1 1 - T 


U - || II' 
12 3 4 5 6 


Step 2 : Calculating the equation of the line 


The equation of the line is 


y = rax + c 


From the graph we have drawn, we estimate the y-intercept to be 1,5. We estimate 


that y = 3,5 when x = 3. So we have that points (3; 3,5) and (0: 1,5) lie on the 


line. The gradient of the line, m, is given by 


2/2 -y\ 


m = 


x 2 — Xi 


3,5-1,5 


3-0 


2 


3 


So we finally have that the equation of the line of best fit is 


2 

y = 3 X + 





The Method of Least Squares 



EMCCY 



We now come to a more accurate method of finding the line of best-fit. The method is very simple. 
Suppose we guess a line of best-fit. Then at at every data point, we find the distance between the 
data point and the line. If the line fitted the data perfectly, this distance should be zero for all the data 
points. The worse the fit, the larger the differences. We then square each of these distances, and add 



l(i<) 



7 7.4 



CHAPTER 11. STATISTICS 



them all together. 



The best-fit line is then the line that minimises the sum of the squared distances. 
Suppose we have a data set of n points {(#1; 3/1), (x 2 ; 2/2), . . . , (x„; y n )}. We also have a line f(x) = 
mx + c that we are trying to fit to the data. The distance between the first data point and the line, for 
example, is 

distance = y\ — /(si) = y\ — (mx\ + c) 

We now square each of these distances and add them together. Lets call this sum S(m,c). Then we 
have that 



S(m,c) 



(Vi - f(xi)f + (V2 - f(x 2 )) 2 + .s + (y n - f{x n )f 



X> - f {x ^ 



Thus our problem is to find the value of m and c such that S(m,c) is minimised. Let us call these 
minimising values m and c a . Then the line of best-fit is f(x) = m x + Co. We can find m and c 
using calculus, but it is tricky, and we will just give you the result, which is that 



mo 



Ca 



i=l xiyi ^LgigLgijij 
n T,7=i( x ') 2 - (E?=i a; 2 

^ n n 

1 ^-^ m -r-^ 

- > yt > Xi=y — m a x 



See video: VMhyr at www.everythingmaths.co.za 



Example 2: Method of Least Squares 



QUESTION 



In the table below, we have the records of the maintenance costs in Hands, compared with 
the age of the appliance in months. We have data for five appliances. 



appliance 


1 


2 


3 


4 


5 


age (x) 


5 


10 


15 


20 


30 


cost (y) 


90 


140 


250 


300 


380 



170 



CHAPTER 11. STATISTICS 



11.4 



SOLUTION 



appliance 


X 


y 


xy 


x 1 


1 


5 


90 


450 


25 


2 


10 


140 


1400 


100 


3 


15 


250 


3750 


225 


4 


20 


300 


6000 


400 


5 


30 


380 


11400 


900 


Total 


80 


1160 


23000 


1650 



:J2xy-J2xJ2y _5x 23000 - 80 x 1160 



nX> 2 -Q» 2 5xl650-i 

1160 12 x 80 



12 



y — bx 



:40 



40 + 12a: 




Example 3: Using the Sharp EL-53 7 VH calculator 



QUESTION 



Find a regression equation for the following data: 



Days (x) 


1 


2 


3 


4 


5 


Growth in m (y) 


1,00 


2,50 


2,75 


3,00 


3,50 



SOLUTION 



Step 1 : Getting your calculator ready 

Using your calculator, change the mode from normal to "Stat xy". This mode 
enables you to type in bivariate data. 



Step 2 : Entering the data 

Key in the data as follows: 



171 



7 7.4 



CHAPTER 11. STATISTICS 



1 


(x,y) 


1 

2,5 
2,75 
3,0 
3,5 


DATA 


n= 1 


2 


(x,y) 


DATA 


n = 2 


3 


(x,y) 


DATA 


n = 3 


4 


{x,y) 


DATA 


n = 4 


5 


(x,y) 


DATA 


n = 5 



Step 3 : Getting regression results from the calculator 

Ask for the values of the regression coefficients a and b. 



RCL 



RCL 



a gives a = 0,9 
6 gives b = 0,55 

.■.£ = 0,9 + 0,55x 



Example 4: Using the CASIO fx-82ES Natural Display calculator 



QUESTION 



Using a calculator determine the least squares line of best fit for the following data set of 
marks. 



Learner 


1 


2 


3 


4 


5 


Chemistry (%) 


52 


55 


86 


71 


45 


Accounting (%) 


48 


64 


95 


79 


50 



For a Chemistry mark of 65%, what mark does the least squares line predict for Account- 



ing^ 



SOLUTION 



Step 1 : Getting your calculator ready 

Switch on the calculator. Press [MODE] and then select STAT by pressing [2]. 
The following screen will appear: 



1 


l-VAR 


2 


A + BX 


3 


-+CX 2 


4 


InX 


5 


eX 


6 


A.B'X 


7 


A.XB 


8 


1/X 



Now press [2] for linear regression. Your screen should look something like this: 



V 



Step 2 : Entering the data 



172 



CHAPTER 11. STATISTICS 



11.4 



Press [52] and then [ = ] to enter the first mark under x. Then enter the other 
values, in the same way, for the x-variable (the Chemistry marks) in the order 
in which they are given in the data set. Then move the cursor across and up 
and enter 48 under y opposite 52 in the x-column. Continue to enter the other 
j/-values (the Accounting marks) in order so that they pair off correctly with the 
corresponding x-values. 



52 
55 



y 



Then press [AC]. The screen clears but the data remains stored. 



1 

3 


Type 
Edit 


2: 
4: 


Data 
Sum 


5 


Var 


6: 


MinMax 


7 


Reg 







Now press [SHIFT][1] to get the stats computations screen shown below. 
Choose Regression by pressing [7]. 



1: 


A 


2: 


B 


3: 


r 


4: 


X 


5: 


y 







Step 3 : Getting regression results from the calculator 

a) Press [1] and [ = ] to get the value of the y-intercept, a = —5,065 . . . = —5,07 (to two 
decimal places) 

Finally, to get the slope, use the following key sequence: [SHIFT][1][7][2][ = ]. The calcu- 
lator gives b = 1,169 . . . = 1,17 (to two decimal places) 

The equation of the line of regression is thus: 
y = -5,07+ 1,17a; 

b) Press [AC][65][SHIFT][1][7][5][ = ] 

This gives a (predicted) Accounting mark of"= 70,94 = 71% 



Exercise 11-3 



1 . The table below lists the exam results for five students in the subjects of Science and Biology. 



Learner 


1 


2 


3 


4 


5 


Science % 


55 


66 


74 


92 


47 


Biology % 


48 


59 


68 


81 


53 



(a) Use the formulae to find the regression equation coefficients a and b. 

(b) Draw a scatter plot of the data on graph paper. 

(c) Now use algebra to find a more accurate equation. 

2. Footlengths and heights of seven students are given in the table below. 



Height (cm) 


170 


163 


131 


181 


146 


134 


166 


Footlength (cm) 


27 


23 


20 


28 


22 


20 


24 



173 



7 7.4 



CHAPTER 11. STATISTICS 



(a) Draw a scatter plot of the data on graph paper. 

(b) Identify and describe any trends shown in the scatter plot. 

(c) Find the equation of the least squares line by using algebraic methods and draw the line on 
your graph. 

(d) Use your equation to predict the height of a student with footlength 21,6 cm. 

(e) Use your equation to predict the footlength of a student 176 cm tall. 

3. Repeat the data in Question 2 and find the regression line using a calculator. 



A" 1 ) More practice f ►) video solutions CfJ or help at www.everythingmaths.co.za 



(1.)01b2 (2.)01b3 (3.)01b4 



Correlation Coefficients 



EMCDA 



Once we have applied regression analysis to a set of data, we would like to have a number that tells 
us exactly how well the data fits the function. A correlation coefficient, r, is a tool that tells us to what 
degree there is a relationship between two sets of data. The correlation coefficient r e [—1; 1] when 
r = — 1, there is a perfect negative correlation, when r = 0, there is no correlation and r = 1 is a 
perfect positive correlation. 



Positive, strong Positive, fairly strong Positive, weak 

r sa 0,9 r « 0,7 r ss 0,4 



No association Negative, fairly strong 

r = rss —0,7 



We often use the correlation coefficient r 2 in order to examine the strength of the correlation only. 
In this case: 



r 2 =0 


no correlation 


< r 2 < 0,25 


very weak 


0,25 < r 2 < 0,5 


weak 


0,5 < r 2 < 0,75 


moderate 


0,75 < r 2 < 0,9 


strong 


0,9 < r 2 < 1 


very strong 


r 2 = l 


perfect correlation 



The correlation coefficient r can be calculated using the formula 

r _ l y^ I ! - ■'' \ / ■" - 'i 



171 



CHAPTER 11. STATISTICS 



11.4 



• where n is the number of data points, 

• s x is the standard deviation of the x-values and 

• s y is the standard deviation of the j/-values. 

This is known as the Pearson's product moment correlation coefficient. It is a long calculation and 
much easier to do on the calculator where you simply follow the procedure for the regression equation, 
and go on to find r. 



Chapter 1 1 



End of Chapter Exercises 



1. Below is a list of data concerning 12 countries and their respective carbon dioxide 
(CO2) emmission levels per person and the gross domestic product (GDP is a measure 
of products produced and services delivered within a country in a year) per person. 





CO2 emmissions per capita (x) 


GDP per capita (y) 


South Africa 


8,1 


3 938 


Thailand 


2,5 


2 712 


Italy 


7,3 


20 943 


Australia 


17.0 


23 893 


China 


2.5 


816 


India 


0,9 


463 


Canada 


16,0 


22 537 


United Kingdom 


9,0 


21785 


United States 


19,9 


31806 


Saudi Arabia 


11,0 


6 853 


Iran 


3,8 


1493 


Indonesia 


1,2 


986 



(a) Draw a scatter plot of the data set and your estimate of a line of best fit. 

(b) Calculate equation of the line of regression using the method of least squares. 

(c) Draw the regression line equation onto the graph. 

(d) Calculate the correlation coefficient r. 

(e) What conclusion can you reach, regarding the relationship between CO2 emis- 
sion and GDP per capita for the countries in the data set? 

2. A collection of data on the peculiar investigation into a foot size and height of stu- 
dents was recorded in the table below. Answer the questions to follow. 



Length of right foot (cm) 


Height (cm) 


25,5 


163,3 


26.1 


164,9 


23,7 


165,5 


26,4 


173,7 


27.5 


174,4 


24 


156 


22,6 


155,3 


27,1 


169,3 



(a) Draw a scatter plot of the data set and your estimate of a line of best fit. 

(b) Calculate equation of the line of regression using the method of least squares or 
your calculator. 

(c) Draw the regression line equation onto the graph. 



175 



7 7.4 



CHAPTER 11. STATISTICS 



(d) Calculate the correlation coefficient r. 

(e) What conclusion can you reach, regarding the relationship between the length 
of the right foot and height of the students in the data set? 

3. A class wrote two tests, and the marks for each were recorded in the table below. Full 
marks in the first test was 50, and the second test was out of 30. 

(a) Is there a strong association between the marks for the first and second test? 
Show why or why not. 

(b) One of the learners (in Row 18) did not write the second test. Given her mark 
for the first test, calculate an expected mark for the second test. 



Learner 


Test 1 
(Full marks: 50) 


Test 2 
(Full marks: 30) 


1 


42 


25 


2 


32 


19 


3 


31 


20 


4 


42 


26 


5 


35 


23 


6 


23 


14 


7 


43 


24 


8 


23 


12 


9 


24 


14 


10 


15 


10 


11 


19 


11 


12 


13 


10 


13 


36 


22 


14 


29 


17 


15 


29 


17 


16 


25 


16 


17 


29 


18 


18 


17 




19 


30 


19 


20 


28 


17 



A fast food company produces hamburgers. The number of hamburgers made, and 
the costs are recorded over a week. 



Hamburgers made 


Costs 


495 


R2 382 


550 


R2 442 


515 


R2 484 


500 


R2 400 


480 


R2 370 


530 


R2 448 


585 


R2 805 



(a) Find the linear regression function that best fits the data. 

(b) If the total cost in a day is R2 500, estimate the number of hamburgers produced. 

(c) What is the cost of 490 hamburgers? 

5. The profits of a new shop are recorded over the first 6 months. The owner wants 
to predict his future sales. The profits so far have been R90 000; R93 000; R99 500; 
R102 000; R101 300; R109 000. 

(a) For the profit data, calculate the linear regression function. 

(b) Give an estimate of the profits for the next two months. 

(c) The owner wants a profit of R130 000. Estimate how many months this will take. 

6. A company produces sweets using a machine which runs for a few hours per day. 
The number of hours running the machine and the number of sweets produced are 
recorded. 



170 



CHAPTER 11. STATISTICS 



11.4 



Machine hours 


Sweets produced 


3,80 


275 


4,23 


287 


4,37 


291 


4,10 


281 


4,17 


286 



Find the linear regression equation for the data, and estimate the machine hours 
needed to make 300 sweets. 



More practice f ►) video solutions C'f) or help at www.everythingmaths.co.: 



(1.)01b5 (2.)01b6 (3.)01b7 (4.)01b8 (5.)01b9 (6.)01ba 



177 



Combinations and 
Permutations 





1 2. 1 Introduction 




Mathematics education began with counting. In the beginning, fingers, beans and buttons were used 
to help with counting, but these are only practical for small numbers. What happens when a large 
number of items must be counted? 

This chapter focuses on how to use mathematical techniques to count combinations of items. 

© See introductory video: VMidw at www.everythingmaths.co.za 




12.2 Counting 




An important aspect of probability theory is the ability to determine the total number of possible 
outcomes when multiple events are considered. 

For example, what is the total number of possible outcomes when a die is rolled and then a coin is 
tossed? The roll of a die has six possible outcomes (1; 2; 3; 4; 5 or 6) and the toss of a coin, 2 outcomes 
(head or tails). Counting the possible outcomes can be tedious. 



Making a List 



EMCDD 



The simplest method of counting the total number of outcomes is by making a list: 

1H; IT; 2H; 2T; 3H; ST; AH; AT; 5H; 5T; 6H; 6T 
or drawing up a table: 



die 


coin 


1 


H 


1 


T 


2 


H 


2 


T 


3 


H 


3 


T 


4 


H 


4 


T 


5 


H 


5 


T 


6 


H 


6 


T 



178 



CHAPTER 12. COMBINATIONS AND PERMUTATIONS 



12.3 



Both these methods result in 12 possible outcomes, but both these methods have a lot of repetition. 
Maybe there is a smarter way to write down the result? 



Tree Diagrams 



EMCDE 



One method of eliminating some of the repetition is to use tree diagrams. Tree diagrams are a graphical 
method of listing all possible combinations of events from a random experiment. 




H T H T H T H T H T H T 



Figure 12.1: Example of a tree diagram. Each possible outcome is a branch of the tree. 




12.3 Notation 




Factorial Notation 



EMCDG 



For an integer n, the notation n! (read n factorial) represents: 

n x (n — 1) x (n — 2) x ■ ■ ■ x 3 x 2 x 1 
with the following definition: 0! = 1. 
The factorial notation will be used often in this chapter. 





12.4 Fundamental Counting 
Principle 



The use of lists, tables and tree diagrams is only feasible for events with a few outcomes. When the 
number of outcomes grows, it is not practical to list the different possibilities and the fundamental 
counting principle is used. 

The fundamental counting principle describes how to determine the total number of outcomes of a 
series of events. 

Suppose that two experiments take place. The first experiment has n\ possible outcomes, and the 
second has n 2 possible outcomes. Therefore, the first experiment, followed by the second experiment, 



179 



12.5 



CHAPTER 12. COMBINATIONS AND PERMUTATIONS 



will have a total of rn x n 2 possible outcomes. This idea can be generalised to m experiments as the 
total number of outcomes for m experiments is: 



n\ X n2 X ri3 X ■ ■ ■ X n„ 



IL 



Yl is the multiplication equivalent of ^2- 

Note: the order in which the experiments are done does not affect the total number of possible out- 
comes. 



Example 1: Lunch Special 



QUESTION 



A take-away has a 4-piece lunch special which consists of a sandwich, soup, dessert and drink 

for R25,00. They offer the following choices for : 

Sandwich: chicken mayonnaise, cheese and tomato, tuna, and ham and lettuce 

Soup: tomato, chicken noodle, vegetable 

Dessert: ice-cream, piece of cake 

Drink: tea, coffee, coke, Fanta and Sprite. 

How many possible meals are there? 



SOLUTION 



Step 1 : Determine how many parts to the meal there are 

There are 4 parts: sandwich, soup, dessert and drink. 

Step 2 : Identify how many choices there are for each part 



Meal component 


Sandwich 


Soup 


Dessert 


Drink 


Number of choices 


4 


3 


2 


5 



Step 3 : Use the fundamental counting principle to determine how many different meals 
are possible 



4x3x2x5 = 120 
So there are 120 possible meals. 




12.5 Combinations 




The fundamental counting principle describes how to calculate the total number of outcomes when 
multiple independent events are performed together. 



1N0 



CHAPTER 12. COMBINATIONS AND PERMUTATIONS 12.5 



A more complex problem is determining how many combinations there are of selecting a group of 
objects from a set. Mathematically, a combination is defined as an un-ordered collection of unique 
elements, or more formally, a subset of a set. For example, suppose you have fifty-two playing cards, 
and select five cards. The five cards would form a combination and would be a subset of the set of 52 
cards. 

In a set, the order of the elements in the set does not matter. These are represented usually with curly 
braces. For example {2; 4; 6} is a subset of the set {1; 2; 3; 4; 5; 6}. Since the order of the elements 
does not matter, only the specific elements are of interest. Therefore, 

{2; 4; 6} = {6; 4; 2} 

and {1; 1; 1} is the same as {1} because in a set the elements don't usually appear more than once. 

So in summary we can say the following: Given S, the set of all possible unique elements, a combina- 
tion is a subset of the elements of S. The order of the elements in a combination is not important (two 
lists with the same elements in different orders are considered to be the same combination). Also, the 
elements cannot be repeated in a combination (every element appears once). 




Calculating the number of ways that certain patterns can be formed is the beginning of combinatorics, 
the study of combinations. Let S be a set with n objects. Combinations of r objects from this set S are 
subsets of S having r elements each (where the order of listing the elements does not distinguish two 
subsets). 



Combination Without Repetition 

When the order does not matter, but each object can be chosen only once, the number of combinations 
is: 



r\(n — r)\ \r 

where n is the number of objects from which you can choose and r is the number to be chosen. 

For example, if you have 10 numbers and wish to choose 5 you would have 10!/(5!(10 — 5)!) = 252 
ways to choose. 

For example how many possible 5 card hands are there in a deck of cards with 52 cards? 

52!/(5!(52 - 5)!) = 2 598 960 combinations 



Combination with Repetition 

When the order does not matter and an object can be chosen more than once, then the number of 
combinations is: 



(ra + r — 1)! In + r — l\ ln + r — 1 



r\(n — 1)! I r J In — 1 

where n is the number of objects from which you can choose and r is the number to be chosen. 

For example, if you have ten types of donuts to choose from and you want three donuts there are 
(10 + 3 — 1)!/3!(10 — 1)! = 220 ways to choose. See video: VMihd at www.everythingmaths.co.za 



1S1 



72.6 



CHAPTER 12. COMBINATIONS AND PERMUTATIONS 



Combinatorics and Probability 



EMCDK 



Combinatorics is quite useful in the computation of probabilities of events, as it can be used to deter- 
mine exactly how many outcomes are possible in a given experiment. 



Example 2: Probability 



QUESTION 



At a school, learners each play 2 sports. They can choose from netball, basketball, soccer, 
athletics, swimming, or tennis. What is the probability that a learner plays soccer and either 
netball, basketball or tennis? 



SOLUTION 



Step 1 : Identify what events we are counting 

We count the events: soccer and netball, soccer and basketball, soccer and 
tennis. This gives three choices. 

Step 2 : Calculate the total number of choices 

There are 6 sports to choose from and we choose 2 sports. There are 
Q =6!/(2!(6-2)!) = 15 choices. 

Step 3 : Calculate the probability 

The probability is the number of events we are counting, divided by the total 
number of choices. 
Probability = ± = | = 0,2 




12.6 Permutations 




The concept of a combination did not consider the order of the elements of the subset to be important. 
A permutation is a combination with the order of a selection from a group being important. For 
example, for the set {1, 2, 3, 4, 5, 6}, the combination {1,2,3} would be identical to the combination 
{3, 2, 1}, but these two combinations are different permutations, because the elements in the set are 
ordered differently. 

More formally, a permutation is an ordered list without repetitions, perhaps missing some elements. 

This means that {1; 2; 2; 3; 4; 5; 6} and {1; 2; 4; 5; 5; 6} are not permutations of the set {1; 2; 3; 4; 5; 6}. 



182 



CHAPTER 12. COMBINATIONS AND PERMUTATIONS 12.6 

Now suppose you have these objects: 

1;2;3 

Here is a list of all permutations of all three objects: 

12 3; 13 2; 2 13; 2 3 1; 3 12; 3 2 1. 



Counting Permutations wemcdm 



Let 5 be a set with n objects. Permutations of r objects from this set S refer to sequences of r different 
elements of S (where two sequences are considered different if they contain the same elements but 
in a different order). Formulae for the number of permutations and combinations are readily available 
and important throughout combinatorics. 

It is easy to count the number of permutations of size r when chosen from a set of size n (with r < n). 

1 . Select the first member of the permutation out of n choices, because there are n distinct elements 
in the set. 

2. Next, since one of the n elements has already been used, the second member of the permutation 
has (n — 1) elements to choose from the remaining set. 

3. The third member of the permutation can be filled in (n — 2) ways since 2 have been used 
already. 

4. This pattern continues until there are r members on the permutation. This means that the last 
member can be filled in (n — (r — 1)) = (n — r + 1) ways. 

5. Summarising, we find that there is a total of 

n(n - l)(n - 2) •• • (n — r + 1) 

different permutations of r objects, taken from a pool of n objects. This number is denoted by 
P(n, r) and can be written in factorial notation as: 

P(n,r) 



(n — r)\ 



For example, if we have a total of 5 elements, the integers {1; 2; 3; 4; 5}, how many ways are there for 
a permutation of three elements to be selected from this set? In this case, n = 5 and r = 3. Then, 
P(5,3) = 5!/7! = 60!. 

© See video: VMihe at www.everythingmaths.co.za 



Example 3: Permutations 



.1 N.->, 



12.7 



CHAPTER 12. COMBINATIONS AND PERMUTATIONS 



QUESTION 



Show that a collection of n objects has n\ permutations. 



SOLUTION 



Proof: Constructing an ordered sequence of n objects is equivalent to choosing the position 
occupied by the first object, then choosing the position of the second object, and so on, until 
we have chosen the position of each of our n objects. 

There are n ways to choose a position for the first object. Once its position is fixed, we 
can choose from (n — 1) possible positions for the second object. With the first two placed, 
there are (n — 2) remaining possible positions for the third object; and so on. There are only 
two positions to choose from for the penultimate object, and the nth object will occupy the 
last remaining position. 

Therefore, according to the fundamental counting principle, there are 

n(n- l)(n-2) ■ ■ ■ 2 x 1 = n\ 
ways of constructing an ordered sequence of n objects. 



Permutation with Repetition 

When order matters and an object can be chosen more than once then the number of 
permutations is: 

n 

where n is the number of objects from which you can choose and r is the number to be chosen. 

For example, if you have the letters A, B, C, and D and you wish to discover the number of ways of 
arranging them in three letter patterns (trigrams) you find that there are 4 3 or 64 ways. This is because 
for the first slot you can choose any of the four values, for the second slot you can choose any of the 
four, and for the final slot you can choose any of the four letters. Multiplying them together gives the 
total. 




12.7 Applications 




Extension: 



The Binomial Theorem 



In mathematics, the binomial theorem is an important formula giving the expansion of powers 
of sums. Its simplest version reads 



/ \n \~^ / n \ k n-k 

(x+ y ) =2^\ k \ x y 

k = ' ' 



181 



CHAPTER 12. COMBINATIONS AND PERMUTATIONS 12.7 



Whenever n is a positive integer, the numbers 



k k\(n-k)\ 

are the binomial coefficients (the coefficients in front of the powers). 
For example, here are the cases n = 2, n = 3 and n = 4: 



(x + %i) = x + Ixy + y 



(x + yf = x 3 + 3x 2 y + 3xy 2 + y 3 

I i \4 4 . ,. 3 ,«22.^ 3, 4 

(a; + y) = x + 4x y + bx y + 4xy + j/ 



The coefficients form a triangle, where each number 
is the sum of the two numbers above it: 



This formula and the triangular arrangement of the binomial coefficients, are often at- 
tributed to Blaise Pascal who described them in the 17th century. It was, however, known 
to the Chinese mathematician Yang Hui in the 1 3th century, the earlier Persian mathematician 
Omar Khayym in the 1 1th century, and the even earlier Indian mathematician Pingala in the 
3rd century BC. 



Example 4: Number Plates 



QUESTION 



The number plate on a car consists of any 3 letters of the alphabet (excluding the vowels and 
'Q'), followed by any 3 digits (0 to 9). For a car chosen at random, what is the probability that 
the number plate starts with a 'Y' and ends with an odd digit? 



SOLUTION 



Step 1 : Identify what events are counted 

The number plate starts with a 'Y', so there is only 1 choice for the first letter, 
and ends with an odd digit, so there are 5 choices for the last digit (1, 3, 5, 7, 9). 

Step 2 : Find the number of events 

Use the counting principle. For each of the other letters, there are 20 possible 
choices (26 in the alphabet, minus 5 vowels and 'Q') and 10 possible choices for 
each of the other digits. 
Number of events = 1 x 20 x 20 x 10 x 10 x 5 = 200 000 

Step 3 : Find the number of total number of possible number plates 

Use the counting principle. This time, the first letter and last digit can be any- 
thing. 
Total number of choices = 20 x 20 x 20 x 10 x 10 x 10 = 8 000 000 

Step 4 : Calculate the probability 

The probability is the number of events we are counting, divided by the total 
number of choices. 



185 



12.7 



CHAPTER 12. COMBINATIONS AND PERMUTATIONS 



Probability - ^§- = £ = 0,025 



Example 5: Factorial 



QUESTION 



Show that 



SOLUTION 



(n-l)\ 



Method 1: Expand the factorial notation. 

n! _ n x (n — 1) x (n — 2) x ■ ■ ■ x 2 x 1 
(n-1)! ~ (n- 1) x (n-2) x ■■■ x 2 x 1 

Cancelling the common factor of (n — 1) x (n — 2) x ■ ■ ■ x 2 x 1 on the top and bottom leaves 

n. 

So T^h = n 

Method 2: We know that P(n,r) = , re " ! ,, is the number of permutations of r objects, 
taken from a pool of n objects. In this case, r = 1. To choose 1 object from n objects, there 
are n choices. 
So 



(n-l)l 



Chapter 12 



End of Chapter Exercises 



1 . Tshepo and Sally go to a restaurant, where the menu is: 



Starter 


Main Course 


Dessert 


Chicken wings 


Beef burger 


Chocolate ice cream 


Mushroom soup 


Chicken burger 


Strawberry ice cream 


Greek salad 


Chicken curry 


Apple crumble 




Lamb curry 


Chocolate mousse 




Vegetable lasagna 





(a) How many different combinations (of starter, main course, and dessert) can 
Tshepo have? 

(b) Sally doesn't like chicken. How many different combinations can she have? 



186 



CHAPTER 12. COMBINATIONS AND PERMUTATIONS 12.7 



2. Four coins are thrown, and the outcomes recorded. How many different ways are 
there of getting three heads? First write out the possibilities, and then use the formula 
for combinations. 

3. The answers in a multiple choice test can be A, B, C, D, or E. In a test of 12 ques- 
tions, how many different ways are there of answering the test? 

4. A girl has 4 dresses, 2 necklaces, and 3 handbags. 

(a) How many different choices of outfit (dress, necklace and handbag) does she 
have? 

(b) She now buys 2 pairs of shoes. How many choices of outfit (dress, necklace, 
handbag and shoes) does she now have? 

5. In a soccer tournament of 9 teams, every team plays every other team. 

(a) How many matches are there in the tournament? 

(b) If there are 5 boys' teams and 4 girls' teams, what is the probability that the first 
match will be played between 2 girls' teams? 

6. The letters of the word "BLUE" are rearranged randomly. How many new words (a 
word is any combination of letters) can be made? 

7. The letters of the word "CHEMISTRY" are arranged randomly to form a new word. 
What is the probability that the word will start and end with a vowel? 

8. There are 2 History classes, 5 Accounting classes, and 4 Mathematics classes at 
school. Luke wants to do all three subjects. How many possible combinations of 
classes are there? 

9. A school netball team has 8 members. How many ways are there to choose a captain, 
vice-captain, and reserve? 

10. A class has 15 boys and 10 girls. A debating team of 4 boys and 6 girls must be 
chosen. How many ways can this be done? 

1 1 . A secret pin number is 3 characters long, and can use any digit (0 to 9) or any letter 
of the alphabet. Repeated characters are allowed. How many possible combinations 
are there? 



(A -1 ) More practice \w\ video solutions (?) or help at www.everythingmaths.co.za 

(1.) Olbb (2.) 01 be (3.)01bd (4.) 01 be (5.)01bf (6.)01bg 

(7.)01bh (8.) Olbi (9.) Olbj (10.)01bk (11.)01bm 



1ST 



VERSION 0.9 NCS 

GRADE 12 
MATHEMATICS 

WRITTEN BY VOLUNTEERS 



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